Question

Determine the point on the curve y=3x^2
-5 at which the normal is parallel to a line whose

slope is−
1/3​

Answer 1

We have to determine the point on the curve y = 3x² - 5 at which the normal is parallel to a line whose slope is -1/3 .

Solution : let a point on the curve is (a, b).

differentiating curve with respect to x,

dy/dx = 6x

at (a, b) , dy/dx = 6a

so slope of tangent = 6a

Then slope of normal = -1/slope of tangent = -1/6a

It is given that, slope of normal is parallel to slope of line whose slope is -1/3

so, -1/6a = -1/3 ⇒a = 1/2

but (a, b) lies on the curve so, b = 3a² - 5

Here, a = 1/2 so, b = 3(1/2)² - 5 = -17/4

therefore point on the curve is (1/2, -17/4)

Answer 2

Answer:

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