Q.464. To find the equation of the in-circle given vertices of a triangle

Question 464.
Find the equation of circle inscribed in the triangle with vertices
(-7,-10),(-7,15) and (5,-1).

Answer 464.
Let A (-7, -10), B (-7, 15) and C (5, -1) be the given vertices of ΔABC

=> a = BC = √[(5+7)^2 + (-1-15)^2] = 20,
b = CA = √[(5+7)^2 + (-1+10)^2] = 15
and c = AB = √(-7+7)^2 + (15+10)^2] = 25

=> coordinates of the in-center are
x = 1/(a+b+c) * (-7a -7b+5c)
= 1/(20+15+25) * (-7*20 - 7*15 + 5*25) = - 2
and y = 1/(a+b+c) * (-10a +15b-c)
= 1/(20+15+25) * (-10*20 +15*15 - 25) = 0
=> incenter = (-2, 0)

Radius of the in-circle
= perpendicular distance from (-2, 0) to the line through AB
= -2 + 7 = 5

=> eqn. of the inscribed circle is
(x + 2)^2 + y^2 = 5^2
=> x^2 + y^2 + 4x - 21 = 0.

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Q.455. Geometry challenge

Question 455.
A circle (radius = r), and an equilateral triangle (side = 2r), fit perfectly in a square, as shown in the diagram.
What is (length CD) divided by (height of triangle)?

Answer 455.

Trigonometric Proof:

Refer to the figure:

Let O be the center of the circle.
With r = 1, (The required ratio is independent of the value of r.)
vertical side of the square
= 1 + ODcos60° + CDcos30°
= 1 + 1/2 + (√3/2) CD
= 3 + (√3/2) CD ... ( 1 )

Horizontal side of the square
= 1 + ODcos30° + (BC - CDcos60°)
= 1 + √3/2 + 2 - CD/2
= 3 + √3/2 - (1/2) CD ... ( 2 )
Equatting ( 1 ) and ( 2 ),
3/2 + (√3/2) CD = 3 + √3/2 - (1/2) CD
=> (√3 - 1)/2 CD = (√3 - 3)/2
=> CD = √3
and Height of the triangle = 2 cos30° = √3
=> CD/Height of the triangle
= √3 / √3
= 1.
=======================================…
Proof using Co-ordinate Geometry:
Refer to the figure:

Consider the given drawn inverted as above.

The required ratio, being independent of the radius, let r = 1
=> The eqn. of the circle with its center as origin is
x^2 + y^2 = 1 ... ( 1 )

Let the length of side of the square = a
=> B = (a-1, a-1), C = (a-3, a-1) and A = (a-2, a-1-√3)
Slope of AC = - √3
Let the eqn. of the tangent AC be y = - √3x + c
=> c = r √(1 + m^2) = 2
=> eqn. of AC is y = - √3x + 2 ... ( 2 )

Solving eqn. ( 1 ) and ( 2 ) gives
x^2 + (-√3x + 2)^2 = 1
=> 4x^2 - 4√3x + 3 = 0
=> (2x - √3) = 0
=> x-coordinate of D is √3/2
Plugging in eqn. ( 2 ),
y-coordinate is 1/2
=> D = (- √3/2, 1/2)

Plugging coordinates of C in eqn. ( 2 ),
a -1 = - √3 (a - 3) + 2
=> a = (3√3 + 3) / (√3 + 1) = 3
=> C = (0, 2)

CD^2 = (0 + √3/2)^2 + (3/2)^2 = 3
=> CD = √3
and Height of the triangle = 2 cos30° = √3
=> CD/Height of the triangle
= √3 / √3
= 1.

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Q.450. To find the equation of a non-standard parabola.

Question 450.
Determine the general equation for the parabola with its focus is located on (6,4) and that its directrix equation is 4x + y -6 = 0.

Answer 450.
Parabola is defined as a set of points equidistant from a given point, called focus and a given line, called directrix.
If P (x, y) is a point on the parabola,
its distance from the focus (6, 4) = √[(x - 6)^2 + (y - 4)^2]
and its distance from the directrix 4x + y - 6 = 0 is
l4x + y - 6l / √(4^2 + 1^2)
=> √[(x - 6)^2 + (y - 4)^2] = l4x + y - 6l / √(4^2 + 1^2)
=> 17 [(x - 6)^2 + (y - 4)^2] = (4x + y - 6)^2
=> 17 (x^2 + y^2 - 12x - 8y + 52) = (16x^2 + y^2 + 36 + 8xy - 12y - 48x)
=> x^2 + 16y^2 - 8xy - 156x - 124y + 848 = 0.

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Q.449. Area of a quadrilateral.

Question 449.
In quadrilateral ABCD it's known that:
AC * BD = x and AB² - BC² + CD² - DA² = y
Find area of quadrilateral in terms of x and y.

Answer 449.
Refer to the following figure.

Let A(0, 0) be at the origin.
Let X-axis be along AC and C = (m, 0)
Let B = (a, b) and D = (c, d)

AC * BD = x
=> m * BD = x
=> BD^2 = x^2/m^2
=> (a - c)^2 + (b - d)^2 = x^2/m^2 ... ( 1 )

AB^2 - BC^2 + CD^2 - DA^2 = y
=> a^2 + b^2 - (m - a)^2 - b^2 + (m - c)^2 + d^2 - c^2 - d^2 = y
=> 2m (a - c) = y

Plugging (a - c) = y/2m in ( 1 ),
y^2/4m^2 + (b - d)^2 = x^2/m^2
=> l b - d l = √[x^2/m^2 - y^2/4m^2]

Area of the quadrilateral
= (1/2) AC * l b - d l
= (1/2) m * √[x^2/m^2 - y^2/4m^2]
= (1/2) √[x^2 - y^2/4)].

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Q.441. Tangents - Co-ordinate geometry.

Question 441.
If  from  any point  on  the  common  chord  of  two  intersecting  circles,  tangents  are  drawn  to  the circles,  prove  that they are of  equal  lengths.

Answer 441.
Let x^2 + y^2 + 2g1x + 2f1y + c1 = 0
and x^2 + y^2 + 2g2x + 2f2y + c2 = 0 be any two circles.
Their common chord is obtained by taking the difference of the equations and is
2(g1 - g2)x + 2(f1 - f2)y + c1 - c2 = 0
Let (h, k) be any point external to the two circles on the above common chord.
=> 2(g1 - g2)h + 2(f1 - f2)k + c1 - c2 = 0 ... ( 1 )

If L1 and L2 are the lengths of tangents from (h, k) to the given circles,
L1^2 = h^2 + k^2 + 2g1h + 2f1k + c1 and
L2^2 = h^2 + k^2 + 2g2h + 2f2k + c2
=> L1^2 - L2^2
= 2(g1 - g2)h + 2(f1 - f2)k + c1 - c2
= 0 ... [from eqn. ( 1 )]
=> L1^2 = L2^2
=> L1 = L2.

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Q.422. To prove an identity related to triangle.

Question 422.
Triangle ABC is right-angled at A. P is the midpoint of AB and Q is the midpoint of BC. Choose suitable coordinates in order to prove that BQ^2 - PC^2 = 3(PB^2 - QC^2).

Answer 422.
Let origin be at A and x-axis be along AB and C on y-axis
=> A = (0, 0), B = (2a, 0) and C = (0, 2b)
=> P = (a, 0) and Q = (a, b)
=> BQ^2 - PC^2
= (2a - a)^2 + (0- b)^2 - [(a - 0)^2 + (0 - 2b)^2]
= - 3b^2 ... ( 1 )
and 3(PB^2 - QC^2)
= 3 [(2a - a)^2 + 0^2 - (a - 0)^2 - (b - 2b)^2]
= - 3b^2 ... ( 2 )
From ( 1 ) and ( 2 ),
BQ^2 - PC^2 = 3(PB^2 - QC^2).

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Q.416. Equation of the ellipse given its vertices and foci.

Question 146.
Find an equation of an ellipse having veticles (-1, 3),  (7, 3), and foci (0, 3), (6,3).

Answer 146.
Midpoint of vertices or foci, (3, 3) is the center of the ellipse.
Shifting origin to (3, 3),
vertices are (-4, 0) and (4, 0) and
foci are (-3, 0) and (3, 0)
=> a = 4 and ae = 3
=> b^2 = a^2 - (ae)^2 = 16 - 9 = 7
=> eqn. of the ellipse with center, (0, 0) is
x^2/16 + y^2/7 = 1
=> eqn. of the ellipse with center (3, 3) is
(x - 3)^2/16 + (y - 3)^2/7 = 1.

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Q.410. Equation of hyperbola given vertices and foci.

Question 410.
Write the equation of a hyperbola with vertices at (1,1) and (9, 1)
and foci at (0, 1) and (10, 1).

Answer 410.
Midpoint of vertices or foci = (5, 1) which is the center of the hyperbola
Shifting origin to (5, 1),
vertices = (-4, 0) and (4, 0) and
foci = (-5, 0) and (5, 0)
=> a = 4 and ae = 5
b^2 = a^2 (e^2 - 1) = 25 - 16 = 9
=> eqn. of the hyperbola with the new origin is
x^2/16 - y^2/9 = 1
=> equation of the hyperbola with the given vertices and foci is
(x-5)^2/16 - (y-1)^2/9 = 1.

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Q.404. Distance of a point from a plane along a direction.

Question 404.
Find the distance of point (1,-2,3) from the plane 2x + 3y - 4z = 16 measured along the straight line with direction ratios 1,-2, 3.

Answer 404.
Point (1, -2, 3) lies on the line
(x, y, z) = (1, -2, 3) + k (1, -2, 3) whose direction ratios are 1, -2, 3 for k = 0.
=> x = 1 + k, y = -2 - 2k and z = 3 + 3k
Plugging in the equation of the plane
2 (1 + k) + 3 (-2 - 2k) - 4 (3 + 3k) = 16
=> - 16 - 16k = 16
=> k = - 2
=> the line intersects the plane in the point (1+k, -2-2k, 3+3k)
= (-1, 2, -3)
=> the required distance is the distance between
(-1, 2, -3) and (1, -2, 3)
= √[(-1 - 1)^2 + (2 + 2)^2 + (-3 -3)^2]
= 2√(14).

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Q.402. Identifying the conic and obtaining its equation.

Question 402.
Obtain the equation of conic , a focus of which lies at (2,1) , the directrix of which is x + y = 0 and which passes through (1,4) . Also identify the conic.

Answer 402.
A conic is a set of points ratio of whose distances from the focus and directrix is
a constant, e, called eccentricity.

If P(x, y) is any point on the conic,
then (x - 2)^2 + (y - 1)^2 = e^2 * (x + y)^2 / 2
As ( 1, 4) lies on it,
(1 - 2)^2 + (4 - 1)^2 = e^2 * (1 + 4)^2 /2
=> 10 = e^2 * 25/2
=> e^2 = 0.8
=> e < 1
=> conic is the ellipse.

To confirm that the conic is the ellipse, see the plot of the curve in the link below.
Wolfram Alpha Graph

 Simplifying, (x - 2)^2 + (y - 1)^2 = e^2 * (x + y)^2 / 2
=> 2x^2 + 2y^2 - 8x - 4y + 10 = (0.80) (x^2 + 2xy + y^2)
=> 1.2 x^2 + 1.2 y^2 - 1.6 xy - 8x - 4y + 10 = 0
=> 3x^2 + 3y^2 - 4xy - 20x - 10y + 25 = 0
is the equation of the required conic.

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Q.398. Equation of a line from polar to Cartesian form

Question 398.
Does this equation represents a straight line ?
The equation : 2/r = 3cos ( θ-π/4) + 2sin (θ + π/4).

Answer 398.
3cos(θ - π/4) + 2sin(θ + π/4)
= 3 [cosθ cos(π/4) + sinθ sin(π/4)] + 2 [sinθ cos(π/4) + cosθ sin(π/4)]
= (5/√2) (cosθ + sinθ)

x = r cosθ and y = rsinθ represents a line with r as a constant
=> cosθ = x/r and sinθ = y/r

Plugging in the given eqn.
2/r = (5/√2) (x/r + y/r)
=> x + y = (2√2)/5
which is the equation of a straight line.

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Q.396. Equation of a plane through a line and perpendicular to a plane

Question 396.
Find the equation of a plane passing through the line
2x-y= 0 = 3z-y and perpendicular to the plane 4x+5y+3z = 8.

Answer 396.
Eqns. of all the planes through the line of intersection of the planes
2x - y = 0 and 3z - y = 0 is
2x - y + k (y - 3z) = 0
=> 2x + (k - 1)y - 3kz = 0
 Let k = k' represent a plane which is perpendicular to the given plane 4x + 5y + 3z = 8
Its normal is n1 = [2, (k'-1), - 3k']
Normal of the given plane 4x + 5y + 3z = 8 is n2 = (4, 5, 3)
=> n1 . n2 = 0
=> 2*4 + (k' - 1)*5 - (3k')*3 = 0
=> 8 + 5k' - 5 - 9k' = 0
=> k' = 3/4
 => eqn. of the reqd. plane is
2x + (3/4 - 1)y - 9z/4 = 0
=> 8x - y - 9z = 0.

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Q.395. Length of normal at an extremity of latus rectum intercepted by parabola

Question 395.
If the length of normal at an extremity of latus rectum of parabola y^2 = 4ax , a > 0 intercepted by parabola is 4 units,
find a.

Answer 395.
Extremity of latus rectum is (a, 2a)
y^2 = 4ax
=> 2y dy/dx = 4a
=> dy/dx = 2a/y
=> slope of normal = dx/dy = - y/2a
=> slope of normal at (a, 2a) = - 1
=> eqn. of normal at (a, 2a) is
y - 2a = - 1 (x - a)
=> y = - x + 3a
Solving with y^2 = 4ax,
(-x + 3a)^2 = 4ax
=> x^2 - 10ax + 9a^2 = 0
=> (x - 9a) (x - a) = 0
=> x = a or 9a
x = 9a => y = - 9a + 3a = - 6a
=> the other point of intersection of normal with parabola is
(9a, - 6a)
=> 4^2 = (9a - a)^2 + (- 6a - 2a)^2
=> 16 = 128 a^2
=> a = 1/(2√2).

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Q.393. Point of intersection of two lines in 3-D.

Question 393.
Find the point of intersection of the lines represented by
(i + 3j) + λ(- i - j) and (i + 2j) + λ(i - j).

Answer 393.
The given lines are
r (vector) = (1, 3) + λ(-1, -1)
r (vector) = (1, 2) + λ(1, -1)
Their directions are (-1, -1) and (1, -1) and the dot product = -1 + 1 = 0
=> they are perpendicular.
Let (x, y) be their point of intersection
=> x = 1 - λ1 = 1 + λ2
and y = 3 - λ1 = 2 - λ2
=> λ1 + λ2 = 0 and λ1 - λ2 = 1
=> λ1 = 1/2 and λ2 = - 1/2

Plugging λ1 = 1/2 in the eqn. of the first line,
(x, y) = (1/2, 5/2)
Plugging λ2 = -1/2 in the eqn. of the second line,
(x, y) = 1/2, 5/2)

=> (1/2, 5/2) is the common point of the two lines and is the point of intersection.

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Q.391. Equation of a normal to a parabola.

Question. 391.
If by = 4x + 5 touches 3y^2 = 8x + 1 at a point P . Find b and coordinates of P . Also find the equation of normal at P .

Answer 391.
Solving by = 4x + 5 with 3y^2 = 8x + 1
=> 3y^2 = 8 * (by - 5)/4 + 1
=> 3y^2 - 2by + 9 = 0
This quadratic shloud have equal roots for the line to be tangent to the parabola
=> discriminant = 0
=> 4b^2 - 108 = 0
=> b = 3√3
y = (1/6) * 2b = √3 and
x = (1/4) (by - 5) = (1/4) * 4 = 1
=> P = (1, √3)

Slope of the line is 4/b = 4/(3√3)
=> slope of the normal is - (3√3)/4
=> equation of the normal through P is
y - √3 = - (3√3)/4 (x - 1)
=> (3√3) x + 4y = 7√3

This answer is verified using the following link.
Wolfram Alpha

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Q. 387. Co-ordinate geometry, Contact points of vertical tangents to a curve

Question 387.
Find the coordinates of all points on the curve (x+y)² + 2(x-y)² =24
at which the tangent is parallel to the y-axis.

Answer 387.
(x + y)^2 + 2(x - y)^2 = 24
=> 3x^2 + 3y^2 - 2xy = 24
=> 6x + 6y dy/dx - 2 (y + x dy/dx) = 0
=> 2 (3x - y) + 2 (3y - x) dy/dx = 0
=> dy/dx = (3x - y)/(x - 3y)

For tangent to be parallel to y-axis, dy/dx is not defined
=> x - 3y = 0
 For point P (x1, y1),
x1 - 3y1 = 0 ... ( 1 )
Also, 3x1^2 + 3y1^2 - 2x1y1 = 24 ... ( 2 )

Plugging x1 = 3y1 from ( 1 ) in ( 2 ),
3 (3y1)^2 + 3y1^2 - 2(3y1) * y1 = 24
=> 24y1^2 = 24
=> y1 = -1 or +1
=> x1 = 3y1 = - 3 or +3
=> points are (-3, -1) and (3, 1).
 
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Q.384. Center and radius of the circle in 3-D.

Question 384.
Find the coordinates of center and radius of the circle of intersection of 
the sphere x²+y²+z²=9 with the plane 3x+4y+5z=5,

Answer 384.
The equation of the family of spheres through the given sphere and the given plane is
x^2 + y^2 + z^2 - 9 + k (3x + 4y + 5k - 5) = 0.

The centre of the intersecting circle is the centre of the sphere from the above family which lies on the plane.
The centre of the sphere = (-3k/2, - 4k/2, - 5k/2)

Plugging in the eqn. of the plane
=> - 9k/2 - 16k/s - 25k/2 - 5 = 0
=> k = -1/5
=> eqn. of the sphere containing the intersecting circle as its large circle is
x^2 + y^2 + z^2 - 9 - (1/5) (3x + 4y + 5z - 5) = 0
Its centre is (3/10, 2/5, 1/2)
Its radius
= √[(3/10)^2 + (4/10)^2 + (5/10)^2 + 9 - 1]
= √[1/2 + 8]
= √(17/2)

Alternate method to find the radius is
radius of the sphere
r = 3
perpendicular distance from the centre (0, 0, 0) of the sphere to the plane
p = 5/√(3^2 + 4^2 + 5^2) = 1/√2
=> radius of the circle of intersection
= √(r^2 - p^2)
= √(9 - 1/2)
= √(17/2).

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Q.381. Angle between tangents drawn from a point to a parabola.

Question 381.
What is the angle between the tangents drawn from the point (1,4) to the parabola y^2 = 4x ?

Answer 381.
y = mx + a/m is the tangent to the parabola y^2 = 4ax
For y^2 = 4x, a = 1
=> y = mx + 1/m is the tangent

If it passes through (1, 4),
4 = m + 1/m
=> m^2 - 4m + 1 = 0
The roots m1 and m2 are the slopes of the tangents
=> m1 + m2 = 4 and m1m2 = 1
=> m1 - m2
= √[(m1 + m2)^2 - 4m1m2]
= √(16 - 4)
= 2√3

If θ = angle between the tangents, then
tanθ
= l (m1 - m2) / (1 + m1m2) l
= l 2√3 / (1 + 1) l
= √3
=> θ = arctan√3 = π/3.

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Q.380. Centroid of a triangle.

Question 380.
Segments AD = 10, BE = 6 and CF = 24 are drawn from the vertices of triangle ABC , each perpendicular to a straight line RS, not intersecting the triangle. Points D, E and F are the intersection points of RS with the perpendiculars. If x is the length of the perpendicular segment, GH, drawn to RS from the intersection point, G, of the medians of the triangle, then find the value of x.

Answer 380.
Treat RS as x-axis.
=> y-coordinate of A, B and C are
AD=10, BE = 6 and CF = 24 respectively.
=> y-coordinate of the median G
= GH = (1/3) (10 + 6 + 24) = 40/3.
=> x = 40/3.

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Q.364. Number of tangents to a curve and points of contacts.

Question 364.
How many tangent lines to the curve y= x/(x + 1) pass through the point (1,2)? At which points do these tangent lines touch the curve?

Answer 364.
The line through (1, 2) having slope m is
y - 2 = m (x - 1)

The tangent to a curve is a line which intersects the curve in two or more identical points. Hence, solving the above line with the curve,
y = x/(x + 1) = 2 + m(x - 1)
=> x = 2(x + 1) + m (x^2 - 1)
=> mx^2 + x + 2 - m = 0
The roots of this quadratic eqn. must be identical
=> its discriminant = 0 and the equal roots are x = 1/(2m)
=> 1 - 4m(2 - m) = 0
=> 4m^2 - 8m + 1 = 0
=> m^2 - 2m + 1/4 = 0
=> (m - 1)^2 = 3/4
=> m - 1 = ±√3/2
=> m = (1/2) (2 ± √3)
Two values of m indicate that there are two tangent lines.

The points of contact are given by
x = - 1/2m = - (2 - √3) or - (2 + √3)
For x = - (2 - √3),
y = x/(x+1) = - (2 - √3)/(-1 + √3) = - (1/2) (2 - √3)*(√3 + 1)   ...   (by rationalizing)
=> y = (1/2) (1 - √3)
Similarly, for x = - (2 + √3), y = (1/2)(1 + √3)
=> the points of intersection are
(-2 ± √3, (1 ∓ √3)/2).

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