The tangent and normal to the ellipse 3x^2+5y^2=32 at the point P(2,2) meet the x-axis at …

Question

The tangent and normal to the ellipse $3x^2+5y^2=32$ at the point $P(2,2)$ meet the $x$-axis at $Q$ and $R$, respectively. Then the area (in sq. units) of the triangle $PQR$ is:

Accepted Answer

$\dfrac{x^2}{32/3}+\dfrac{y^2}{32/5}=1$. Tangent at $P$: $\dfrac{3x}{16}+\dfrac{5y}{16}=1$; at $y=0$: $Q=\left(\dfrac{16}3,0\right)$.

Normal at $P$: $\dfrac{32x}{3(2)}-\dfrac{32y}{5(2)}=\dfrac{32}3-\dfrac{32}5$; at $y=0$: $R=\left(\dfrac45,0\right)$.

$$PQ=\left|2-\frac{16}3\right|=\frac{10}3,\qquad PR=\left|2-\frac45\right|=\frac65,\qquad QR=\frac{16}3-\frac45=\frac{68}{15}.$$

Since $Q,R$ are on the $x$-axis and $P$ has height $2$:
$$\text{Area}=\frac12\cdot QR\cdot2=\frac{68}{15}.$$

Solutions · 1