Question

72
4.
A tangent having slope of
to the ellipse e ti = 1 intersects the major & minor axes in
3
points A & B respectively. If C is the centre of the ellipse, then the area of the triangle ABC is
(A) 12 sq. units
(B) 24 sq. units (C) 36 sq. units (D) 48 sq. units​

Answer 1

Option (B) 24 sq.units.

Given : A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$ intersects the major and minor axes in three points A, B and C respectively. If C is the centre of the ellipse, then the area of the ΔABC is : ?

To find : Area of the ΔABC.

Solution : From the given data we have slope and values of a and b.

Slope (m) =  $-\frac{4}{3}$

a = 18 ; b = 32

General equation :    $y = mx + \sqrt{a^{2}m^2+b^{2} }$

Now substituting the respective values, it gives

                                  $y = \frac{-4}{3} x+\sqrt{(18)^{2}\times(-\frac{4}{3})^2+(32)^2 }$                                  

                                  $y =- \frac{4}{3}x+\sqrt{(18)\times (\frac{16}{9})+ (32)}$

                                  $y =- \frac{4}{3}x+\sqrt{ 32+ 32}$

                                  $y =- \frac{4}{3}x+\sqrt{64}$

                                  $y =- \frac{4}{3}x+8$   -----> (1)

To find the value of x and y :

To find the value of x :

Put y = 0 in equation (1)

      $y =- \frac{4}{3}x+8$

      $0 =- \frac{4}{3}x+8$

      $- \frac{4}{3}x= -8$

           $x=\frac{8\times3}{4}$

           $x=\frac{24}{4}$

           $x = 6$

Hence, the point is A(x, y) = (6, 0).

To find the value of y :

Put x = 0 in equation (1)

     $y =- \frac{4}{3}\times0+8$

     $y =0+8$

     $y = 8$

So, the point is B(x,y) = (0, 8).

Area of the triangle ABC = $\frac{1}{2}\times\{base\}\times\{height\}$

Base = 6  ;  Height = 8

Area of ΔABC = $\frac{1}{2}\times\{6\}\times\{8\}$

                       = $\frac{1}{2}\times\{48\}$

                       = 24 sq.units

Therefore, the area of ΔABC is 24 sq.units.

To learn more...

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