Question

If $P$ and $Q$ are points on ellipse (s) $\frac{x^2}{9}+\frac{y^2}{16}=2$ such that $\triangle PQR$ is an isosceles triangle with $PR=QR$ and its area equal to 6 sq. units, where $R=(0,-6)$, then $P$ can have the co-ordinates

• A. $(3, -4)$
• B. $(-3, -4)$
• C. $(3, 4)$
• D. $(3\sqrt{2}, 0)$

Answer 1

Answer: (A), (B)

$(3, -4)$ and $(-3, -4)$

Answer 2

The given ellipse equation is $\frac{x^2}{9}+\frac{y^2}{16}=2$, which can be rewritten in standard form as $\frac{x^2}{18}+\frac{y^2}{32}=1$.

The condition $PR=QR$ implies that P and Q must be symmetric with respect to the perpendicular bisector of PQ, which passes through R(0,-6). Assuming P and Q are symmetric with respect to the y-axis ($x_Q = -x_P$, $y_Q = y_P$), the area of $\triangle PQR$ with R=(0,-6) is given by $6 = \frac{1}{2} |x_P(y_P - (-6)) - x_Q(-6 - y_P)|$. Substituting $x_Q = -x_P$, we get $12 = |x_P(y_P+6) - (-x_P)(-6-y_P)| = |x_P(y_P+6) + x_P(y_P+6)| = |2x_P(y_P+6)|$. Thus, $|x_P(y_P+6)| = 6$.

We need to find points $(x,y)$ on the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$ that satisfy $|x(y+6)| = 6$.

Let's check the option $(3, -4)$:

1. Is it on the ellipse? $\frac{3^2}{18} + \frac{(-4)^2}{32} = \frac{9}{18} + \frac{16}{32} = \frac{1}{2} + \frac{1}{2} = 1$. Yes.
2. Does it satisfy $|x(y+6)| = 6$? $|3(-4+6)| = |3(2)| = |6| = 6$. Yes.

Let's check the option $(-3, -4)$:

1. Is it on the ellipse? $\frac{(-3)^2}{18} + \frac{(-4)^2}{32} = \frac{9}{18} + \frac{16}{32} = \frac{1}{2} + \frac{1}{2} = 1$. Yes.
2. Does it satisfy $|x(y+6)| = 6$? $|-3(-4+6)| = |-3(2)| = |-6| = 6$. Yes.

Let's check the option $(3, 4)$:

1. Is it on the ellipse? $\frac{3^2}{18} + \frac{4^2}{32} = \frac{9}{18} + \frac{16}{32} = \frac{1}{2} + \frac{1}{2} = 1$. Yes.
2. Does it satisfy $|x(y+6)| = 6$? $|3(4+6)| = |3(10)| = |30| = 30$. No.

Let's check the option $(3\sqrt{2}, 0)$:

1. Is it on the ellipse? $\frac{(3\sqrt{2})^2}{18} + \frac{0^2}{32} = \frac{18}{18} + 0 = 1$. Yes.
2. Does it satisfy $|x(y+6)| = 6$? $|3\sqrt{2}(0+6)| = |18\sqrt{2}| \neq 6$. No.

Therefore, the coordinates of P can be $(3, -4)$ and $(-3, -4)$.