Question

the line 4y=x+c, where c is a constant, is a tangent to the curve y^2=x+3 at the point P on the curve

Answer 1

SOLUTION

GIVEN

The line 4y = x + c, where c is a constant, is a tangent to the curve y² = x + 3 at the point P on the curve

TO DETERMINE

• The value of c

• The point P

EVALUATION

Let the coordinates of the point P is (h, k)

Now the given equation of the line is

4y = x + c

Slope of the line is

$\displaystyle \sf{ \frac{1}{4} }$

Again the given equation of the curve is

$\displaystyle \sf{ {y}^{2} = x + 3}$

Differentiating both sides with respect to x we get

$\displaystyle \sf{2y \: \frac{dy}{dx} = 1 }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{1}{2y} }$

Hence slope of the line at the point (h, k) is

$\displaystyle \sf{ \frac{1}{2k} }$

So by the given condition

$\displaystyle \sf{ \frac{1}{2k} = \frac{1}{4} }$

$\displaystyle \sf{ \implies \: k = 2 }$

Again (h, k) is a point on the curve y² = x + 3

$\displaystyle \sf{ \implies \: {k}^{2} = h + 3 }$

$\displaystyle \sf{ \implies \: {2}^{2} = h + 3 }$

$\displaystyle \sf{ \implies \:h = 1 }$

Hence the required point is P(1,2)

Again (h, k) is a point on the line 4y = x + c

$\displaystyle \sf{ \implies \: 4k = h + c }$

$\displaystyle \sf{ \implies \: c = 8 - 1}$

$\displaystyle \sf{ \implies \: c = 7 }$

Hence the required value of c = 7

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