Question

The locus of a point which moves such that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distances from the other two sides is

• A. Hyperbola
• B. A parabola
• C. An ellipse
• D. A ciircle

Answer 1

The correct option is A Hyperbola

If the triangle is PQR, with $PQ=PR$, take P, Q, R to be the points $(0,b),(–a,0),(a,0)$ respectively.

The equation of the line PQ is

$y-b=\frac{b}{+a}\left(x\right)$

$ay-ab=bx$

$bx-ay+ab=0$

and the equation of PR is

$y-b=\frac{b}{-a}\left(x\right)$

$-ay+ab=bx$

$bx+ay-ab=0$

and the equation of QR is

$y=\frac{0}{2a}\left(x\right)$

$y=0$

To find the locus of a point $X(h,k)$ which moves so that the square of its distance from QR is equal to the product of its distances from $PQ$ and $PR$. The distance from $X(h,k)$ to PQ is

$d_1=\left|\frac{bh-ak+ab}{\sqrt{a^2+b^2}}\right|$

and the distance from $X(h,k)$ to PR is

$d_2=\left|\frac{bh+ak-ab}{\sqrt{a^2+b^2}}\right|$

The distance from $X(h,k)$ to QR is $|k|$

So according to question

The distance of $X$ to QR=product of distances from $X$ to PQ and PR

$k=d_1{d}_2$

$k=\frac{bh-(ak-ab)}{\sqrt{a^2+b^2}}\times \frac{bh+(ak-ab)}{\sqrt{a^2+b^2}}$

$k=\frac{b^2{h}^2-(a^2{k}^2+a^2{b}^2-2a^{2}kb)}{a^2+b^2}$

$a^{2}k+b^{2}k=b^2{h}^2-a^2{k}^2-a^2{b}^2+2a^{2}bk$

Putting $h=x,k=y$

$b^2{x}^2+(2a^2+b^2)y^2+2a^{2}by-a^2{b}^2=0$

Hence above equation represents the pair of straight lines