Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on conics. Click on this to access more such competitive quizzes on conic sections.

Conics Quiz 2

mathematics

 A given triangle is formed by the three lines $x-2y+3=0$, $4x+5y-40=0$ and y-axis. Find the equation of the circle that can be drawn on the line AC as its diameter.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/40638695-33e4-4025-8f8d-c9e3a86cc375241052898458259403.png"/>

 Find the coordinates of point C and then find the equation of the circle using two endpoints of diameter i.e., A and C.

  $\text{To find out the point C which intersect lines AC and BC.} \\ x-2y+3=0 \\ 4x+5y-40=0 \\ \Rightarrow x=5,y=4 \\\therefore \text{Required equation of a circle }=(x-{{x}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}}) \\ \text{ where }({{x}_{1}},{{y}_{1}})=(0,8)\text{ and }({{x}_{2,}}{{y}_{2}})=(5,4)\\ \Rightarrow (x-0)(x-5)+(y-8)(y-4)=0 \\ \Rightarrow {{x}^{2}}+{{y}^{2}}-5x-12y+32=0$ 

 Which of the following coordinates of foci are satisfying the ellipse given by $64{{x}^{2}}+16{{y}^{2}}=9$? 

 Convert the equation in the form of $\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1$ to find $a$ and $b$. Then find $c$ using formula.

  $\text{Given: }64{{x}^{2}}+16{{y}^{2}}=9\\ \Rightarrow \dfrac{64{{x}^{2}}}{9}+\dfrac{16{{y}^{2}}}{9}=1 \\ \Rightarrow \dfrac{{{x}^{2}}}{{{\left( \dfrac{3}{8} \right)}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( \dfrac{3}{4} \right)}^{2}}}=1 \\ \text{Since }\dfrac{3}{4}>\dfrac{3}{8}\text{ so the major axis is along the y-axis} \\ \text{The coordinates of foci are }(0,\pm c). \\ \Rightarrow c=\sqrt{{{a}^{2}}-{{b}^{2}}},\text{ where }a=\dfrac{3}{4}\text{  and }b=\dfrac{3}{8}\\ \Rightarrow c=\sqrt{\dfrac{9}{16}-\dfrac{9}{64}} \\ \Rightarrow c=\dfrac{3\sqrt{3}}{8}$ 

 If the sum of the length of the major axis and minor axis of an ellipse ${{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}$ is 16 and the product of the semi major and semi minor axis is 12, then find the coordinates of foci.

 Using the sum and product condition find the value of $a – b$. Then, find $c$ by simply putting the value of $(a + b)$ and $(a – b)$ in the formula.

 $\text{Let the coordinates of the foci of an ellipse }{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}} \\ \text{  i.e.,  }\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\text{ are }(\pm c,0). \\ \text{Length of (Major axis + Minor axis) }= 2a+2b=16 \\ \Rightarrow a+b=8 \\ \text{Also, ab=12}\\ \Rightarrow a-b=\sqrt{{{(a+b)}^{2}}-4ab} \\ \Rightarrow a-b=\sqrt{64-48}=4 \\ \therefore c=\sqrt{{{a}^{2}}-{{b}^{2}}}=\sqrt{(a+b)(a-b)} \\ \Rightarrow c=4\sqrt{2}$ 

 If the vertices and the length of the latus rectum of the ellipse are $(\pm 7,0)$ and $\dfrac{32}{7}$, then find the eccentricity of the ellipse?

 Using the value of $a$ and the latus rectum find ${{b}^{2}}$. Then, solve for eccentricity by using formula.

 $\text{Given: a = 7 and Length of latus rectum }=\dfrac{2{{b}^{2}}}{a}\\ \Rightarrow \dfrac{2{{b}^{2}}}{a}=\dfrac{32}{7} \\ \Rightarrow {{b}^{2}}=16\text{           }(\because a=7) \\ \therefore e=\dfrac{c}{a} \\ \Rightarrow e=\sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}} \\ \Rightarrow e=\sqrt{\dfrac{49-16}{49}}=\dfrac{\sqrt{33}}{7}$ 

 Which of the following will be the eccentricity of the conic section given by $y=\dfrac{3}{2}\left( {{t}^{2}}+\dfrac{1}{{{t}^{2}}} \right)$ and $x=\dfrac{3}{4}\left( {{t}^{2}}-\dfrac{1}{{{t}^{2}}} \right)$ 

 First find the form of conic section and then find the eccentricity using formula.

 $\text{Given: }y=\dfrac{3}{2}\left( {{t}^{2}}+\dfrac{1}{{{t}^{2}}} \right)\text{ and }x=\dfrac{3}{4}\left( {{t}^{2}}-\dfrac{1}{{{t}^{2}}} \right)\\ \Rightarrow {{x}^{2}}=\dfrac{9}{16}{{\left( {{t}^{2}}-\dfrac{1}{{{t}^{2}}} \right)}^{2}}\text{and  }{{y}^{2}}=\dfrac{9}{4}{{\left( {{t}^{2}}+\dfrac{1}{{{t}^{2}}} \right)}^{2}} \\ \Rightarrow \dfrac{16{{x}^{2}}}{9}+2={{t}^{4}}+\dfrac{1}{{{t}^{4}}}\text{ and }\dfrac{4{{y}^{2}}}{9}-2={{t}^{4}}+\dfrac{1}{{{t}^{4}}} \\ \Rightarrow \dfrac{16{{x}^{2}}}{9}+2=\dfrac{4{{y}^{2}}}{9}-2 \\ \Rightarrow \dfrac{{{y}^{2}}}{{{3}^{2}}}-\dfrac{{{x}^{2}}}{{{\left( \dfrac{3}{2} \right)}^{2}}}=1 \\ \text{This is the standard form of hyperbola.} \\ e=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}=\dfrac{\sqrt{9+\dfrac{9}{4}}}{3} \\ \Rightarrow e=\dfrac{\sqrt{5}}{2}$ 