Question

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then ,the mid-point of the line segment MN must lie on the curve

• A. $4(x+y{)}^2=3xy$
• B. $c^2(x^2+y^2)=4x^2{y}^2$
  
  
• C. $(x^2+y{)}^2=2xy$
• D. $(x^2+y{)}^2=x^2{y}^2$

Answer 1

The correct option is B

$c^2(x^2+y^2)=4x^2{y}^2$

There seems to be insufficient data in question or error in the multiple choices provided. Please review the below solution and make changes to questions as needed.

Refer attached image.

Let the circle be of form:

$x^2+y^2=c^2$

Since S lies at the circle,

$a^2+b^2=c^2$ --- Eqn (1)

Now, tangent at S

$T:xa+by=c^2$

$\Rightarrow \frac{x}{\frac{c^2}{a}}+\frac{y}{\frac{c^2}{b}}=1$

$\Rightarrow M(\frac{c^2}{a},0)\&N(0,\frac{c^2}{b})$

Thus mid-point will be

$R(\frac{c^2}{2a},\frac{c^2}{2b})$

Let this point be (h,k). Thus:

$(h,k)=(\frac{c^2}{2a},\frac{c^2}{2b})$

$\Rightarrow h=\frac{c^2}{2a}\&k=\frac{c^2}{2b}$

$\Rightarrow a=\frac{c^2}{2h}\&b=\frac{c^2}{2k}$

Putting this in Eqn (1), we get

$\frac{c^4}{4h^2}+\frac{c^4}{4k^2}=c^2$

$\Rightarrow c^4(h^2+k^2)=4c^2{h}^2{k}^2$

$\Rightarrow c^2(h^2+k^2)=4h^2{k}^2$

Thus, locus

$c^2(x^2+y^2)=4x^2{y}^2$

No option matches. Maybe due to insufficient data in the question.