120552
The length of the latus rectum of an ellipse is \(\frac{1}{3}\) of the major axis. It's eccentricity is
1 \(\frac{2}{3}\)
2 \(\sqrt{\frac{2}{3}}\)
3 \(\frac{5.4 .3}{7^3}\)
4 \(\left(\frac{3}{4}\right)^4\)
Explanation:
B Given, latus rectum, \(l=\frac{1}{3}\)
According to question,
\(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{1}{3}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{3}\)
Now, \(\mathrm{e}^2=1-\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\mathrm{e}=\sqrt{\frac{3-1}{3}}\)
\(\mathrm{e}=\sqrt{\frac{2}{3}}\)
EAMCET-1991
Ellipse
120553
If the latus rectum of an ellipse subtends a right angle at the center of that ellipse, then the eccentricity of that ellipse is
1 \(\frac{\sqrt{5}+1}{4}\)
2 \(\frac{\sqrt{5}-1}{2}\)
3 \(\frac{\sqrt{10-2 \sqrt{5}}}{5}\)
4 \(\frac{\sqrt{10+2 \sqrt{5}}}{5}\)
Explanation:
B According to question -
\(\mathrm{h}=\mathrm{ae}, \mathrm{p}=\mathrm{q}=\frac{\mathrm{b}^2}{\mathrm{a}}\)
Where, \(\mathrm{h}=\) height of triangle
\(\mathrm{p}=\mathrm{q} \text {, sides of triangle (hypotanous) }\)
\(\therefore \quad \mathrm{a}^2 \mathrm{e}^2 =\frac{\mathrm{b}^4}{\mathrm{a}^2}\)
\(\mathrm{e}^2 =\frac{\mathrm{b}^4}{\mathrm{a}^4}\)
\(\mathrm{e} =\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
Also, \(\quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\mathrm{e}}\)
\(\mathrm{e}=\frac{-1 \pm \sqrt{1+4}}{2}\)\(\mathrm{e}=\frac{\sqrt{5}-1}{2} \quad\) (only positive part)
AP EAMCET-24.04.2018
Ellipse
120554
The mid-point of a chord of the ellipse \(x^2+4 y^2-2 x+20 y=0\) is \((2,-4)\), The equation of the chord is
1 \(x-6 y=26\)
2 \(x+6 y=26\)
3 \(6 x-y=26\)
4 \(6 x+y=26\)
Explanation:
A Given,
\(x^2+4 y^2-2 x+20 y=0 \text { point }(2,-4)\)
\((x-1)^2+4\left(y+\frac{5}{2}\right)^2=26\)
\(\frac{(x-1)^2}{26}+\frac{\left(y+\frac{5}{2}\right)^2}{\frac{26}{4}}=1\)
Equation of chord of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\) with mid point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}+\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=\frac{\mathrm{x}_1^2}{\mathrm{a}^2}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}\)
So,
\(\frac{(x-1)\left(x_1-1\right)}{26}+\frac{\left(y+\frac{5}{2}\right)\left(y_1+\frac{5}{2}\right)}{\frac{26}{4}}=\frac{\left(x_1-1\right)^2}{26}+\frac{\left(y_1+\frac{5}{2}\right)^2}{\frac{26}{4}}\)
\(\frac{x-1}{26}+\frac{4 \times\left(\frac{-3}{2}\right)\left(y+\frac{5}{2}\right)}{26}=\frac{1}{26}+\frac{9}{26}\)
\((x-1)-6\left(y+\frac{5}{2}\right)=10\)
\(\quad x-1-6 y-15=10\)
\(x-6 y=26\)
120552
The length of the latus rectum of an ellipse is \(\frac{1}{3}\) of the major axis. It's eccentricity is
1 \(\frac{2}{3}\)
2 \(\sqrt{\frac{2}{3}}\)
3 \(\frac{5.4 .3}{7^3}\)
4 \(\left(\frac{3}{4}\right)^4\)
Explanation:
B Given, latus rectum, \(l=\frac{1}{3}\)
According to question,
\(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{1}{3}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{3}\)
Now, \(\mathrm{e}^2=1-\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\mathrm{e}=\sqrt{\frac{3-1}{3}}\)
\(\mathrm{e}=\sqrt{\frac{2}{3}}\)
EAMCET-1991
Ellipse
120553
If the latus rectum of an ellipse subtends a right angle at the center of that ellipse, then the eccentricity of that ellipse is
1 \(\frac{\sqrt{5}+1}{4}\)
2 \(\frac{\sqrt{5}-1}{2}\)
3 \(\frac{\sqrt{10-2 \sqrt{5}}}{5}\)
4 \(\frac{\sqrt{10+2 \sqrt{5}}}{5}\)
Explanation:
B According to question -
\(\mathrm{h}=\mathrm{ae}, \mathrm{p}=\mathrm{q}=\frac{\mathrm{b}^2}{\mathrm{a}}\)
Where, \(\mathrm{h}=\) height of triangle
\(\mathrm{p}=\mathrm{q} \text {, sides of triangle (hypotanous) }\)
\(\therefore \quad \mathrm{a}^2 \mathrm{e}^2 =\frac{\mathrm{b}^4}{\mathrm{a}^2}\)
\(\mathrm{e}^2 =\frac{\mathrm{b}^4}{\mathrm{a}^4}\)
\(\mathrm{e} =\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
Also, \(\quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\mathrm{e}}\)
\(\mathrm{e}=\frac{-1 \pm \sqrt{1+4}}{2}\)\(\mathrm{e}=\frac{\sqrt{5}-1}{2} \quad\) (only positive part)
AP EAMCET-24.04.2018
Ellipse
120554
The mid-point of a chord of the ellipse \(x^2+4 y^2-2 x+20 y=0\) is \((2,-4)\), The equation of the chord is
1 \(x-6 y=26\)
2 \(x+6 y=26\)
3 \(6 x-y=26\)
4 \(6 x+y=26\)
Explanation:
A Given,
\(x^2+4 y^2-2 x+20 y=0 \text { point }(2,-4)\)
\((x-1)^2+4\left(y+\frac{5}{2}\right)^2=26\)
\(\frac{(x-1)^2}{26}+\frac{\left(y+\frac{5}{2}\right)^2}{\frac{26}{4}}=1\)
Equation of chord of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\) with mid point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}+\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=\frac{\mathrm{x}_1^2}{\mathrm{a}^2}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}\)
So,
\(\frac{(x-1)\left(x_1-1\right)}{26}+\frac{\left(y+\frac{5}{2}\right)\left(y_1+\frac{5}{2}\right)}{\frac{26}{4}}=\frac{\left(x_1-1\right)^2}{26}+\frac{\left(y_1+\frac{5}{2}\right)^2}{\frac{26}{4}}\)
\(\frac{x-1}{26}+\frac{4 \times\left(\frac{-3}{2}\right)\left(y+\frac{5}{2}\right)}{26}=\frac{1}{26}+\frac{9}{26}\)
\((x-1)-6\left(y+\frac{5}{2}\right)=10\)
\(\quad x-1-6 y-15=10\)
\(x-6 y=26\)
120552
The length of the latus rectum of an ellipse is \(\frac{1}{3}\) of the major axis. It's eccentricity is
1 \(\frac{2}{3}\)
2 \(\sqrt{\frac{2}{3}}\)
3 \(\frac{5.4 .3}{7^3}\)
4 \(\left(\frac{3}{4}\right)^4\)
Explanation:
B Given, latus rectum, \(l=\frac{1}{3}\)
According to question,
\(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{1}{3}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{3}\)
Now, \(\mathrm{e}^2=1-\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\mathrm{e}=\sqrt{\frac{3-1}{3}}\)
\(\mathrm{e}=\sqrt{\frac{2}{3}}\)
EAMCET-1991
Ellipse
120553
If the latus rectum of an ellipse subtends a right angle at the center of that ellipse, then the eccentricity of that ellipse is
1 \(\frac{\sqrt{5}+1}{4}\)
2 \(\frac{\sqrt{5}-1}{2}\)
3 \(\frac{\sqrt{10-2 \sqrt{5}}}{5}\)
4 \(\frac{\sqrt{10+2 \sqrt{5}}}{5}\)
Explanation:
B According to question -
\(\mathrm{h}=\mathrm{ae}, \mathrm{p}=\mathrm{q}=\frac{\mathrm{b}^2}{\mathrm{a}}\)
Where, \(\mathrm{h}=\) height of triangle
\(\mathrm{p}=\mathrm{q} \text {, sides of triangle (hypotanous) }\)
\(\therefore \quad \mathrm{a}^2 \mathrm{e}^2 =\frac{\mathrm{b}^4}{\mathrm{a}^2}\)
\(\mathrm{e}^2 =\frac{\mathrm{b}^4}{\mathrm{a}^4}\)
\(\mathrm{e} =\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
Also, \(\quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\mathrm{e}}\)
\(\mathrm{e}=\frac{-1 \pm \sqrt{1+4}}{2}\)\(\mathrm{e}=\frac{\sqrt{5}-1}{2} \quad\) (only positive part)
AP EAMCET-24.04.2018
Ellipse
120554
The mid-point of a chord of the ellipse \(x^2+4 y^2-2 x+20 y=0\) is \((2,-4)\), The equation of the chord is
1 \(x-6 y=26\)
2 \(x+6 y=26\)
3 \(6 x-y=26\)
4 \(6 x+y=26\)
Explanation:
A Given,
\(x^2+4 y^2-2 x+20 y=0 \text { point }(2,-4)\)
\((x-1)^2+4\left(y+\frac{5}{2}\right)^2=26\)
\(\frac{(x-1)^2}{26}+\frac{\left(y+\frac{5}{2}\right)^2}{\frac{26}{4}}=1\)
Equation of chord of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\) with mid point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}+\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=\frac{\mathrm{x}_1^2}{\mathrm{a}^2}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}\)
So,
\(\frac{(x-1)\left(x_1-1\right)}{26}+\frac{\left(y+\frac{5}{2}\right)\left(y_1+\frac{5}{2}\right)}{\frac{26}{4}}=\frac{\left(x_1-1\right)^2}{26}+\frac{\left(y_1+\frac{5}{2}\right)^2}{\frac{26}{4}}\)
\(\frac{x-1}{26}+\frac{4 \times\left(\frac{-3}{2}\right)\left(y+\frac{5}{2}\right)}{26}=\frac{1}{26}+\frac{9}{26}\)
\((x-1)-6\left(y+\frac{5}{2}\right)=10\)
\(\quad x-1-6 y-15=10\)
\(x-6 y=26\)
120552
The length of the latus rectum of an ellipse is \(\frac{1}{3}\) of the major axis. It's eccentricity is
1 \(\frac{2}{3}\)
2 \(\sqrt{\frac{2}{3}}\)
3 \(\frac{5.4 .3}{7^3}\)
4 \(\left(\frac{3}{4}\right)^4\)
Explanation:
B Given, latus rectum, \(l=\frac{1}{3}\)
According to question,
\(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{1}{3}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{3}\)
Now, \(\mathrm{e}^2=1-\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\mathrm{e}=\sqrt{\frac{3-1}{3}}\)
\(\mathrm{e}=\sqrt{\frac{2}{3}}\)
EAMCET-1991
Ellipse
120553
If the latus rectum of an ellipse subtends a right angle at the center of that ellipse, then the eccentricity of that ellipse is
1 \(\frac{\sqrt{5}+1}{4}\)
2 \(\frac{\sqrt{5}-1}{2}\)
3 \(\frac{\sqrt{10-2 \sqrt{5}}}{5}\)
4 \(\frac{\sqrt{10+2 \sqrt{5}}}{5}\)
Explanation:
B According to question -
\(\mathrm{h}=\mathrm{ae}, \mathrm{p}=\mathrm{q}=\frac{\mathrm{b}^2}{\mathrm{a}}\)
Where, \(\mathrm{h}=\) height of triangle
\(\mathrm{p}=\mathrm{q} \text {, sides of triangle (hypotanous) }\)
\(\therefore \quad \mathrm{a}^2 \mathrm{e}^2 =\frac{\mathrm{b}^4}{\mathrm{a}^2}\)
\(\mathrm{e}^2 =\frac{\mathrm{b}^4}{\mathrm{a}^4}\)
\(\mathrm{e} =\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
Also, \(\quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\mathrm{e}}\)
\(\mathrm{e}=\frac{-1 \pm \sqrt{1+4}}{2}\)\(\mathrm{e}=\frac{\sqrt{5}-1}{2} \quad\) (only positive part)
AP EAMCET-24.04.2018
Ellipse
120554
The mid-point of a chord of the ellipse \(x^2+4 y^2-2 x+20 y=0\) is \((2,-4)\), The equation of the chord is
1 \(x-6 y=26\)
2 \(x+6 y=26\)
3 \(6 x-y=26\)
4 \(6 x+y=26\)
Explanation:
A Given,
\(x^2+4 y^2-2 x+20 y=0 \text { point }(2,-4)\)
\((x-1)^2+4\left(y+\frac{5}{2}\right)^2=26\)
\(\frac{(x-1)^2}{26}+\frac{\left(y+\frac{5}{2}\right)^2}{\frac{26}{4}}=1\)
Equation of chord of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\) with mid point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}+\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=\frac{\mathrm{x}_1^2}{\mathrm{a}^2}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}\)
So,
\(\frac{(x-1)\left(x_1-1\right)}{26}+\frac{\left(y+\frac{5}{2}\right)\left(y_1+\frac{5}{2}\right)}{\frac{26}{4}}=\frac{\left(x_1-1\right)^2}{26}+\frac{\left(y_1+\frac{5}{2}\right)^2}{\frac{26}{4}}\)
\(\frac{x-1}{26}+\frac{4 \times\left(\frac{-3}{2}\right)\left(y+\frac{5}{2}\right)}{26}=\frac{1}{26}+\frac{9}{26}\)
\((x-1)-6\left(y+\frac{5}{2}\right)=10\)
\(\quad x-1-6 y-15=10\)
\(x-6 y=26\)
