Question

$A$: Pole of the line $21x-6y=12$ with respect to the ellipse $3x^2+4y^2=12$
$B$: The positive vertex of $x^2+3y^2=12$
$C$: $C$entre of the ellipse $\frac{(x-2{)}^2}{16}+\frac{(y+3{)}^2}{25}=1$
$D$: End point of Latus rectum of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. ( In the first quadrant) Write the points of the above statements in the decending order of their abscissae.

• A. $A,B,D,C$
• B. $A,D,B,C$
• C. $A,C,D,B$
• D. $B,C,D,A$

Answer 1

The correct option is A $A,B,D,C$

We solve for all the given options.

A: line is $21x-6y=12$ and ellipse is $3x^2+4y^2=12$

The equation of the polar w.r.t. the ellipse is given by $3x{x}_1+4y{y}_1-12=0$

Compare it with the given equation of line to get,

$\frac{3x_1}{21}=\frac{4y_1}{-6}=\frac{-12}{-12}\Rightarrow \frac{x_1}{7}=\frac{2y_1}{-3}=1$

Solve to get $x_1=7,y_1=\frac{-3}{2}$

So abscissa is $7$.

B. Given ellipse $x^2+3y^2=12$ or $\frac{{\left(x\right)}^2}{{\left(2\sqrt{3}\right)}^2}+\frac{{\left(y\right)}^2}{4}=1$

So we see that $a=2\sqrt{3}$

So vertex is $(a,0)\Rightarrow (2\sqrt{3},0)$

So abscissa is $2\sqrt{3}$ .

C. $\frac{{(x-2)}^2}{16}+\frac{{(y+3)}^2}{25}=1$

we can see that the center is $(2,-3)$ So the abscissa is $2$.

D. $\frac{{\left(x\right)}^2}{16}+\frac{{\left(y\right)}^2}{9}=1$

End point of latus rectum in 1st quadrant will be given by $(ae,\frac{b^2}{a})$

We see that $a=4,b=3$ so eccentricity $e=\frac{\sqrt{7}}{4}$

So now end point of latus rectum comes out to be $(\sqrt{7},\frac{9}{4})$

So abscissa is $\sqrt{7}$

So descending order of abscissae is $A,B,D,C$