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  • The tangent and the normal lines at the point to the circle<img alt="4 x^2+2 y^2=4" src="https://ent

The tangent and the normal lines at the point\sqrt{7,2} to the circle4 x^2+2 y^2=4  and the X-axis forms a triangle. The area of this triangle (in square units) is

 

Option: 1

1.65


Option: 2

3.30


Option: 3

4.95


Option: 4

6.60


Answers (1)

best_answer

The equation of the circle centered at the origin by dividing both sides of the given equation by 4:

x^2 / 1+y^2 / 2=1

So the radius of the circle is\sqrt{(2)} and its center is at the origin.

The tangent line to the circle at the point is(\sqrt{(} 7), 2) perpendicular to the radius of the circle at that point. 

So the slope of the tangent line is the negative reciprocal of the slope of the radius at that point.

The slope of the radius is(2-0) /(\sqrt{(} 7)-0)=2 / \sqrt{(} 7) 

so the slope of the tangent line is-\sqrt{(} 7) / 2

Using point-slope form, the equation of the tangent line isy-2=(-\sqrt{(} 7) / 2)(x-\sqrt{(} 7))$ or $y=(-\sqrt{(} 7) / 2) x+(3 \sqrt{(} 7) / 2)

 Using point-slope form, the equation of the normal line is

y-2=(2 / \sqrt{(} 7))(x-\sqrt{(} 7))$ or $y=(2 / \sqrt{(7)}) x-(4 / \sqrt{(7)})

The tangent line and the normal line intersect with the x-axis, we substitute y=0 into each equation and solve for x:\begin{aligned} & (-\sqrt{(} 7) / 2) x+(3 \sqrt{(} 7) / 2)=0 \Rightarrow x=3 \\ & (2 / \sqrt{(} 7)) x-(4 / \sqrt{(} 7))=0 \Rightarrow x=2 \sqrt{(} 7) \end{aligned}

So the vertices of the triangle are (\sqrt{(} 7) / 2),(3,0) and (2 \sqrt{(}(7), 0)To find the base of the triangle, we calculate the distance between (3,0) and (2 \sqrt{(} 7), 0) : distance =|3-2 \sqrt{(7)}|=1.65

Therefore, the area of the triangle formed by the tangent and normal lines at the point 

\left(\sqrt{(7), 2)} \text { to the circle } 4 x^2+2 y^2=4 \text { and the } x \text {-axis is } 1.65\right. \text { square units. }

 

Posted by

seema garhwal

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