Ellipse

Given, equation: 9x2+4y2=72
Differentiate w.r.t x, we get
‌18x+8y⋅‌

dy

dx

=0
⇒‌‌‌

dy

dx

=‌

−18x

8y

=‌

−9x

4y

Slope of tangent at point (2,3) is
mt=‌

−9(2)

4(3)

=‌

−18

12

=‌

−3

2

So, equation of tangent using point-slope form is
‌(y−3)‌=−‌

3

2

(x−2)
⇒‌2y−6‌=−3x+6
⇒‌3x+2y‌=12
Now, slope of normal, mn=‌

−1

mt

=‌

2

3

Now, slope of normal, mn=‌

−1

mt

=‌

2

3

So, equation of normal is
‌(y−3)‌=‌

2

3

(x−2)
‌⇒3y−9‌=2x−4
‌⇒2x−3y‌=−5
Now, for the x-intercept of the tangent line, put y=0, then
‌‌‌3x+2(0)‌=12
⇒‌‌x‌=4
‌‌ so ‌(4,0)
For the x-intercept of the normal line, put y=0, then
‌2x−3(0)=−5
⇒x=−‌

5

2

,‌ so, ‌(‌

−5

2

,0)
Now, base of triangle = Distance b∕wx-intercepts of the tangent and normal lines
b=4−(‌

−5

2

)=‌

13

2

And, height of triangles is the y-coordinate of the point (2,3) is h=3
So, area =‌

1

2

bh=‌

1

2

(‌

13

2

)×3=‌

39

4

∴‌ Area ‌=‌

39

4

‌ sq. units ‌