First slide

Tangents and normals

Question

A point on $y = {(x - 3)}^{2}$ , where the tangent is parallel to the chord joining (3, 0) and (4, 1) is:

Easy

A

$(\frac{7}{2}, \frac{1}{2})$
B

$(\frac{7}{2}, \frac{1}{4})$
C

$(1, 4)$
D

$(4, 1)$

Solution

Let the point be $(x_{1}, y_{1})$ . Therefore, $y_{1} = {(x_{1} - 3)}^{2}$
Now, slope at the tangent at $(x_{1}, y_{1})$ is $2 (x_{1} - 3)$ but it is equal to1.

Therefore $2 (x_{1} - 3) = 1$

$\Rightarrow x_{1} = \frac{7}{2}$

$∴ y_{1} = {(\frac{7}{2} - 3)}^{2} = \frac{1}{4}$

Hence, the point is $(\frac{7}{2}, \frac{1}{4})$

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