Question

L is the circle with equation x2 + y2 = 4
P(3/2, root 7/2) is a point on L.

Find an equation of the tangent to L at the point P.
(3 marks)

Answer 1

The equation of the tangent to L at the point  p(3/2, $\sqrt$7/2) will be equal to  y+$\sqrt$7x/3=0

What is tangent?

A straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.

It is given that the equation of a circle is given by:

$x^2+y^2=4$

Slope formula: If a line passes through two points, then the slope of the line is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

The endpoints of the radius are O(0,0) and P( 3/2,$\sqrt$7/2). So, the slope of the radius is

Now by putting the values in the formula:

$m=\dfrac{\dfrac{3}{2}-0}{\dfrac{\sqrt{7}}{2}-0}$

$m=\dfrac{3}{\sqrt{7}}$

Now the Product of slopes of two perpendicular lines is always -1.

Let the slope of tangent line l is m. Then, the product of slopes of line l and radius is -1.

$m\times m_1=-1$

$m_1\times \dfrac {3}{\sqrt{7}}=-1$

$m_1=\dfrac{-\sqrt{7}}{3}}$

The slope of line l is -$\sqrt$7/3 and it passes through point P(3/2,$\sqrt$7/2). So, the equation of line l is

$y-y_2=m(x-x_2)$

$y-\dfrac{\sqrt{7}}{2}=\dfrac{-\sqrt{7}}{3}(x-\dfrac{3}{2})$

$y-\dfrac{\sqrt{7}}{2}=\dfrac{-\sqrt{7}}{3}x-\dfrac{-\sqrt{7}}{3} \times \dfrac{3}{2})$

$y=\dfrac{-\sqrt{7}}{3}x$

Hence the tangent to L at the point P will have a slope of  $y=\dfrac{-\sqrt{7}}{3}x$

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