Question

If a quadrilateral formed by four tangents to the ellipse 3x + 4y = 12 is a square then
(A) The vertices of the square lie on y = x
(B) The vertices of the square lie on x^2 + y^2 = 7
(C) The area of all such squares is constant
(D) Only two such squares are possible​

Answer 1

Answer:

according to me the correct optipn is (b)

Answer 2

Please correct your equation of ellipse. According to the question the equation of ellipse is $3x^{2} +4y^{2} = 12$

Answer:

The correct option will be (B).

Step-by-step explanation:

The equation of ellipse is  $3x^{2} + 4y^{2} =12$ (given)

We can write it as, $\frac{x^{2} }{4} + \frac{y^{2} }{3} = 1$,

from the above equation, $a^{2} = 4 , b^{2} = 3$

We know that, the equation of tangent from the ellipse is

 $y = mx$ ± $\sqrt{a^{2} m^{2}+b^{2} }$,

where, m = slope of tangents.

From the figure, slope of tangent, m = ±1

After putting all the values in the equation of tangent, we get

So, y = ± $x$  ±$\sqrt{(3+4)}$

      $y=$ ± $x$ ±$\sqrt{7}$

these four tangents are formed a square, whose vertices are $(\sqrt{7} , 0), (-\sqrt{7} , 0), (0, \sqrt{7} ) and ( 0, -\sqrt{7} )$.

According to options, option (B) is correct.

Given, The equation of circle is $x^{2} + y^{2} = 7$

 Radius = $\sqrt{7}$

So, all the vertices of square lies on this circle.