If the sum of the slopes of the normal from a point `P` to the hyperbola `x y=c^2` is equal to `lambda(lambda in R^+)` , then the locus of point `P` is (a) `x^2=lambdac^2` (b) `y^2=lambdac^2` (c) `x y=lambdac^2` (d) none of these

Answer

`x^(2)=lambdac^(2)`

Explanation

Equation of hyperbola is `xy = c^(2)`.
Differentiating w.r.t. x, we get
`(dy)/(dx)=-(y)/(x)`
Thus, slope of normal at any point on the hyperbola is `-(dx)/(dy)=(x)/(y)`.
If point on the hyperbola is (x, y)`-=` (ct, c/t), then `-(dx)/(dy)=t^(2)`.
So, equation of normal at point (ct, c/t) is
`y-(c)/(t)=t^(2)(x-ct)`
`rArr" "ct^(4)=xt^(3)+yt-c=0`
If this normal passes through the point P(h, k) on the plane, then we have
`ct^(4)-ht^(3)+kt-c=0" (1)"`
This is polynomial equation of four degree in variable t.
So, we can get maximum four real roots of the equation.
Hence, maximum four normals can be drawn from point `p(h,k)`. Now, given that sum of the slopes of normals is `lambda`.
i.e., `t_(1)^(2)+t_(2)^(2)+t_(3)^(2)+t_(4)^(2)=lambda`, where `t_(1),t_(2),t_(3) and t_(4)` are roots of equation (1).
`therefore" "(t_(1)+t_(2)+t_(3)+t_(4))^(2)-sumt_(1)t_(2)=0`
`rArr" "((h)/(c))^(2)-0=lambda`
[From equation (1), using sum of roots and sum of product of roots taking two at a time]
`rArr" "h^(2)=c^(2)lambda`
Therefore, required locus is `x^(2)=lambdac^(2).`