The equation of the tangent to L at the point p(3/2, [tex]\sqrt[/tex]7/2) will be equal to y+[tex]\sqrt[/tex]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:
[tex]x^2+y^2=4[/tex]
Slope formula: If a line passes through two points, then the slope of the line is
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
The endpoints of the radius are O(0,0) and P( 3/2,[tex]\sqrt[/tex]7/2). So, the slope of the radius is
Now by putting the values in the formula:
[tex]m=\dfrac{\dfrac{3}{2}-0}{\dfrac{\sqrt{7}}{2}-0}[/tex]
[tex]m=\dfrac{3}{\sqrt{7}}[/tex]
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.
[tex]m\times m_1=-1[/tex]
[tex]m_1\times \dfrac {3}{\sqrt{7}}=-1[/tex]
[tex]m_1=\dfrac{-\sqrt{7}}{3}}[/tex]
The slope of line l is -[tex]\sqrt[/tex]7/3 and it passes through point P(3/2,[tex]\sqrt[/tex]7/2). So, the equation of line l is
[tex]y-y_2=m(x-x_2)[/tex]
[tex]y-\dfrac{\sqrt{7}}{2}=\dfrac{-\sqrt{7}}{3}(x-\dfrac{3}{2})[/tex]
[tex]y-\dfrac{\sqrt{7}}{2}=\dfrac{-\sqrt{7}}{3}x-\dfrac{-\sqrt{7}}{3} \times \dfrac{3}{2})[/tex]
[tex]y=\dfrac{-\sqrt{7}}{3}x[/tex]
Hence the tangent to L at the point P will have a slope of [tex]y=\dfrac{-\sqrt{7}}{3}x[/tex]
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