Question

An ellipse has the points $(1,-1)$ and $(2,-1)$ as its foci and $x+y-5=0$ as one of its tangents. Then the point where this line touches the ellipse from origin is

• A. $(\frac{32}{9},\frac{22}{9})$
• B. $(\frac{23}{9},\frac{2}{9})$
• C. $(\frac{34}{9},\frac{11}{9})$
• D. None of these

Answer 1

The correct option is C $(\frac{34}{9},\frac{11}{9})$

Equation of tangent is $x+y-5=0$.

Foci of ellipse are $(1,-1)$ and $(2,-1)$.

Product of perpendicular from foci upon any tangent is equal to square of the semi minor axis.

Thus $\left|\frac{1-1-5}{\sqrt{1+1}}\right|\left|\frac{2-1-5}{\sqrt{1+1}}\right|=b^2$

$\Rightarrow b^2=\frac{5\times 4}{2}=10$

Distance between foci $=2ae=\sqrt{{(1-2)}^2+{(-1-(-1\left)\right)}^2}$

$\Rightarrow ae=\frac{1}{2}\Rightarrow a^2{e}^2=\frac{1}{4}\Rightarrow a^2(1-\frac{b^2}{a^2})=\frac{1}{4}\Rightarrow a^2-b^2=\frac{1}{4}\Rightarrow a^2-10=\frac{1}{4}\Rightarrow a^2=\frac{41}{4}$

Centre of the ellipse is the mid point of foci.

Therefore, center is $(\frac{3}{2},-1)$.

Therefore, the equation of ellipse is

$\frac{{(x-\frac{3}{2})}^2}{\left(\frac{41}{4}\right)}+\frac{{(y+1)}^2}{10}=1\frac{(2x-3{)}^2}{41}+\frac{{(y+1)}^2}{10}=1$

Substituting $x$ from the equation of tangent, we have

$\Rightarrow \frac{\left(2\right(5-y)-3{)}^2}{41}+\frac{{(y+1)}^2}{10}=1\Rightarrow \frac{(7-2y{)}^2}{41}+\frac{{(y+1)}^2}{10}=1\Rightarrow 10\{49+4y^2-28y\}+41\{y^2+1+2y\}=410$

$\Rightarrow 490+40y^2-280y+41y^2+41+82y=410$

$\Rightarrow 81y^2-198y+121=0$

${\Rightarrow (9y-11)}^2=0\Rightarrow y=\frac{11}{9}$

Now, $x+y-5=0$

$\Rightarrow x=5-y\Rightarrow x=5-\frac{11}{9}\Rightarrow x=\frac{34}{9}$

So, the point of contact of tangent to ellipse is $(\frac{34}{9},\frac{11}{9})$.

Ellipse also passes through this point.

So, option C is correct.