If the normal to the ellipse 3x^2+4y^2=12 at a point P on it is parallel to the line 2x+y=…

Question

If the normal to the ellipse $3x^2+4y^2=12$ at a point $P$ on it is parallel to the line $2x+y=4$ and the tangent to the ellipse at $P$ passes through $Q(4,4)$, then $PQ$ is equal to:

Accepted Answer

$\dfrac{x^2}4+\dfrac{y^2}3=1$. Normal at $(x_1,y_1)$: $\dfrac{4x}{x_1}-\dfrac{3y}{y_1}=1$, slope $=\dfrac{4y_1}{3x_1}=-2\Rightarrow y_1=-\dfrac32x_1$.

On the ellipse: $3x_1^2+4\left(\dfrac94x_1^2\right)=12\Rightarrow12x_1^2=12\Rightarrow x_1=\pm1$.

Tangent: $3xx_1+4yy_1=12$; testing $(x_1,y_1)=(-1,\frac32)$ through $Q(4,4)$: $3(4)(-1)+4(4)(\frac32)=-12+24=12$ .

$$P=\left(-1,\frac32\right),\qquad PQ=\sqrt{5^2+2.5^2}=\sqrt{31.25}=\frac{5\sqrt5}2.$$

Solutions · 1