# Find the equation of the locus of the point of intersection of two tangents drawn to the hyperbola x^2/7-y^2/5=1 such that the sum of the cubes of their slopes is 8. - Mathematics and Statistics

Find the equation of the locus of the point of intersection of two tangents drawn to the hyperbola x^2/7-y^2/5=1 such that the sum of the cubes of their slopes is 8.

#### Solution

Let  p(x_1,y_1)  be a point on the locus. The equation of the hyperbola is x^2/7-y^2/5=1 for which a^2=7, b^2=5

The equation of the tangents to the hyperbola x^2/a^2-y^2/b^2 =1 in terms of slope m are

y=mx+-sqrt(a^2m^2-b^2)

the equation of the tangents to the given hyperbola are

y=mx+-sqrt(7m^2-5)

If these tangents pass through the point p(x_1,y_1)  we get

y_1=mx_1+-sqrt(7m^2-5)

y

y_1-mx_1=+-sqrt(7m^2-5)

On squaring both sides, we get,

(y_1-mx_1)^2=7m^2-5

y_1^2-2mx_1y_1+m^2x_1^2=7m^2-5

(x_1^2-7)m^2-2mx_1y_1+(y_1^2+5)=0

The roots m_1,m_2 of this quadratic equation in m are the slopes of the tangents drawn from the point P to the
hyperbola.

{ (m_1+m_2=(2x_1y_1)/(x_1^2-7)),(and m_1m_2=(y_1^2+5)/(x_1^2-7)) :} ...........(1)

We are given that,
sum of the cubes of the slopes = 8

m_1^3+m_2^3=8

(m_1+m_2)^3-3m_1m_2(m_1+m_2)=8

((2x_1y_1)/(x_1^2-7))^3-3((y_1^2+5)/(x_1^2-7))((2x_1y_1)/(x_1^2-7))=8 .................by(1)

8x_1^3y_1^3-6x_1y_1(y_1^2+5)(x_1^2-7)=8(x_1^2-7)^3

8x_1^3y_1^3-6x_1y_1(y_1^2+5)(x_1^2-7)-8(x_1^2-7)^3=0

the equation of the locus of  P(x_1,y_1) is

8x^3y^3-6xy(y_1^2+5)(x_1^2-7)-8(x_1^2-7)^3=0

Concept: Conics - Locus of Points from Which Two Tangents Are Mutually Perpendicular
Is there an error in this question or solution?
2012-2013 (March)

Share