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 =‌