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Find the locus of the foot of perpendic...

Find the locus of the foot of perpendicular from the centre upon any normal to line hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`.

A

`(x^(2)-+y^(2))^(2)((a^(2))/(x^(2))+(b^(2))/(y^(2))) =(a^(2)-b^(2))^(2)`

B

`(x^(2)+y^(2))^(2)((a^(2))/(x^(2))-(b^(2))/(y^(2)))=(a^(2)+b^(2))^(2)`

C

`(x^(2)+y^(2))^(2) ((x^(2))/(a^(2))-(y^(2))/(b^(2))) =(a^(2)+b^(2))^(2)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
Let the foot of perpendicular from center upon any normal be `P(h,k)` Then equation of normal to hyperbola is
`(y-k) =- (h)/(k) (x-h)`
or `hx + ky = h^(2) +k^(2)` (1)
Also normal at any point `R(a sec theta, tan theta)` on the hyperbola is
`(ax)/(sec theta) +(by)/(tan theta) = a^(2) + b^(2)` (2)
Comparing ratio coefficients of equations (1) and (2),
We get `((a)/(sec theta))/(h)=((b)/(tan theta))/(k) =(a^(2)+b^(2))/(h^(2)+k^(2))`
`rArr sec theta = (a(h^(2)+k^(2)))/(h(a^(2)+b^(2)))` and `tan theta = (b(h^(2)+k^(2)))/(k(a^(2)+b^(2)))`
Squaring and subtracting `(x^(2)+y^(2))^(2) ((a^(2))/(x^(2))-(b^(2))/(y^(2))) = (a^(2)+b^(2))^(2)`
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