Question

Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$. The orthocentre of the $△PAB$ is

• A. $(5,\frac{8}{7})$
• B. $\left(\frac{7}{5},\frac{25}{8}\right)$
• C. $\left(\frac{11}{5},\frac{8}{5}\right)$
• D. $\left(\frac{8}{25},\frac{7}{5}\right)$

Answer 1

The correct option is C $\left(\frac{11}{5},\frac{8}{5}\right)$

In the given problem, we have to draw the chord of contact AB

and find its equation, which is described by the equation $T=0$

Here, P(3,4) and from equation of ellipse, we write

$\frac{3x}{9}+\frac{4y}{4}=1\Rightarrow \frac{x}{3}+\frac{y}{1}=1$

$x+3y=3$ $⟶\left(i\right)$

Now we have to determine the point of intersection of this chord and ellipse

$\frac{x^2}{9}+\frac{y^2}{4}=1$ $⟶\left(ii\right)$

From eq (i) and (ii)

$\Rightarrow \frac{(3-3y{)}^2}{9}+\frac{y^2}{4}=1\Rightarrow \frac{9(1-y{)}^2}{9}+\frac{y^2}{4}=1$

$\Rightarrow \frac{(1-y{)}^2}{1}+\frac{y^2}{4}=1$

$\Rightarrow (1+y^2-2y)+\frac{y^2}{4}=1$

$\Rightarrow \frac{5y^2}{4}-2y=0$

$y=0,\frac{8}{5}$

Putting this in (i), we get $x=3,\frac{-9}{5}$

Equation of line perpendicular to AB is

$3x-y=c⟶\left(iii\right)$

Altitude will pass through Point P (3, 4)

so, put the value of x and y from point P in Eq (iii) we get

$c=5$

Hence

$3x-y=5⟶\left(iv\right)$

Therefore we know Point A (3,0) and P (3,4), thus we will find equation of PA passing through these points,

which will be x = 3, Equation of line perpendicular to PA and passing through B ($\frac{-9}{5},\frac{8}{5}$) will be

$y=\frac{8}{5}⟶\left(v\right)$

(iv) and (v) are altitudes of $\Delta PAB$

Thus, their point of intersection gives orthocenter : $(\frac{11}{5},\frac{8}{5})$