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.
From this quadratic equation,
`{ (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`
