InterviewSolution
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InterviewSolution
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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`x^2=lambdac^2`(b) `y^2=lambdac^2``x y=lambdac^2`(d) none of these |
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Answer» Correct Answer - `x^(2)=lambdac^(2)` 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).` |
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