JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Ellipse

The equation of a tangent to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with slope $m$ is:


Since the tangent passes through $P(3,4)$:



Squaring both sides:



So one tangent is $y=\frac{1}{2}x+\sqrt{\frac{9}{4}+4}=\frac{1}{2}x+\frac{5}{2}$, or $x-2y+5=0$.

To find the point of tangency, use the condition that for a tangent at $(x_0,y_0)$ on the ellipse:


Comparing with $x-2y+5=0$:


From the third equation, $k=5$. Thus $x_0=\frac{9}{5}$, but this doesn't match. Let me recalculate.

Actually, the chord of contact from $P(3,4)$ is:


The points $A$ and $B$ lie on both the ellipse and this line. Substituting $x=3-3y$ into the ellipse equation:







For $y=0$: $x=3$, so $A=(3,0)$.

For $y=\frac{8}{5}$: $x=3-3\cdot \frac{8}{5}=3-\frac{24}{5}=-\frac{9}{5}$, so $B=\left(-\frac{9}{5},\frac{8}{5}\right)$.