120495
The length of the latus rectum of \(3 x^2-4 y+6 x-3=0\) is
1 \(\frac{3}{4}\)
2 \(\frac{4}{3}\)
3 2
4 3
Explanation:
B Given, length of the latus rectum of
\(3 x^2-4 y+6 x-3=0\)
\(3 x^2+6 x-3=4 y\)
\(3\left(x^2+2 x-1\right)=4 y\)
\(3\left(x^2+2 x+1-2\right)=4 y\)
\(3(x+1)^2-6=4 y\)
\(3(x+1)^2=4 y+6\)
\((x+1)^2=\frac{4}{3}(y+3 / 2)\)
Let, \(\quad x^2=\frac{4}{3} y\)
Where, \(\mathrm{X}=\mathrm{x}+1\) and \(\mathrm{Y}=\mathrm{y}+\frac{3}{2}\)
And \(\quad 4 \mathrm{~b}=\frac{4}{3}\)
\(\mathrm{b}=\frac{1}{3}\)So, the length of latusrectum is \(\frac{4}{3}\)
Karnataka CET-2011
Ellipse
120496
Eccentricity of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) if it passes through point \((9,5)\) and \((12,4)\) is
1 \(\sqrt{3 / 4}\)
2 \(\sqrt{4 / 5}\)
3 \(\sqrt{5 / 6}\)
4 \(\sqrt{6 / 7}\)
Explanation:
D We have \(\frac{81}{\mathrm{a}^2}+\frac{25}{\mathrm{~b}^2}=1\)
\(\frac{144}{\mathrm{a}^2}+\frac{16}{\mathrm{~b}^2}=1\)
From equation (ii) - equation (i),
\(\frac{63}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=0 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{7}\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1-\frac{1}{7}}=\sqrt{\frac{6}{7}}\)
BITSAT-2018
Ellipse
120497
If \(\frac{x}{m a}+\frac{y}{n b}=1\) touches the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), then
1 \(\mathrm{m}^2=\frac{\mathrm{n}^2}{\mathrm{n}^2-1}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2-1}\)
2 \(\mathrm{m}^2=\frac{\mathrm{n}^2}{\mathrm{n}^2+1}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2+1}\)
3 \(\mathrm{m}^2=\frac{\mathrm{n}^2+1}{\mathrm{n}^2}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2+1}{\mathrm{~m}^2}\)
4 \(\mathrm{m}^2=\frac{\mathrm{n}^2-1}{\mathrm{n}^2}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2-1}{\mathrm{~m}^2}\)
Explanation:
A Given that,
\(\frac{\mathrm{x}}{\mathrm{ma}}+\frac{\mathrm{y}}{\mathrm{nb}}=1\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1,\)
The line is \(y=-\frac{n b}{m a} x-n b\).
It will touch the ellipse if,
\((-n b)^2=a^2\left(-\frac{n b}{m a}\right)^2+b^2 {\left[c^2=a^2 m^2+b^2\right]}\)
\(n^2=\frac{n^2}{m^2}+1\)
\(m^2=\frac{n^2}{n^2-1} \text { or } \mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2-1}\)
BITSAT-2020
Ellipse
120498
If the area of the ellipse is \(\frac{x^2}{25}+\frac{y^2}{\lambda^2}=1\) is \(20 \pi\) square units, then \(\lambda\) is
1 \(\pm 4\)
2 \(\pm 3\)
3 \(\pm 2\).
4 \(\pm 1\)
Explanation:
(B) Given that,
Equation of focal chord, \(x=2 y+3\),
Eccentricity \(=\frac{3}{4}\)
Foci \(( \pm \mathrm{ae}, 0)\)
Since, \(x=2 y+3\) is a focal chord.
\(\therefore\) It passes through foci.
ae \(=2(0)+3\)
\(\mathrm{a}=\frac{3}{\mathrm{e}}=\frac{3 \times 4}{3}=4\)
Now, \(b^2=a^2\left(1-e^2\right)\)
\(\mathrm{b}^2=(4)^2\left(1-\left(\frac{3}{4}\right)^2\right)=7\)
\(\mathrm{b}= \pm \sqrt{7}\)
\(\therefore\) Length of major axis, \(2 \mathrm{a}=8\)
And length of minor axis, \(2 b=2 \sqrt{7}\)
120495
The length of the latus rectum of \(3 x^2-4 y+6 x-3=0\) is
1 \(\frac{3}{4}\)
2 \(\frac{4}{3}\)
3 2
4 3
Explanation:
B Given, length of the latus rectum of
\(3 x^2-4 y+6 x-3=0\)
\(3 x^2+6 x-3=4 y\)
\(3\left(x^2+2 x-1\right)=4 y\)
\(3\left(x^2+2 x+1-2\right)=4 y\)
\(3(x+1)^2-6=4 y\)
\(3(x+1)^2=4 y+6\)
\((x+1)^2=\frac{4}{3}(y+3 / 2)\)
Let, \(\quad x^2=\frac{4}{3} y\)
Where, \(\mathrm{X}=\mathrm{x}+1\) and \(\mathrm{Y}=\mathrm{y}+\frac{3}{2}\)
And \(\quad 4 \mathrm{~b}=\frac{4}{3}\)
\(\mathrm{b}=\frac{1}{3}\)So, the length of latusrectum is \(\frac{4}{3}\)
Karnataka CET-2011
Ellipse
120496
Eccentricity of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) if it passes through point \((9,5)\) and \((12,4)\) is
1 \(\sqrt{3 / 4}\)
2 \(\sqrt{4 / 5}\)
3 \(\sqrt{5 / 6}\)
4 \(\sqrt{6 / 7}\)
Explanation:
D We have \(\frac{81}{\mathrm{a}^2}+\frac{25}{\mathrm{~b}^2}=1\)
\(\frac{144}{\mathrm{a}^2}+\frac{16}{\mathrm{~b}^2}=1\)
From equation (ii) - equation (i),
\(\frac{63}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=0 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{7}\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1-\frac{1}{7}}=\sqrt{\frac{6}{7}}\)
BITSAT-2018
Ellipse
120497
If \(\frac{x}{m a}+\frac{y}{n b}=1\) touches the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), then
1 \(\mathrm{m}^2=\frac{\mathrm{n}^2}{\mathrm{n}^2-1}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2-1}\)
2 \(\mathrm{m}^2=\frac{\mathrm{n}^2}{\mathrm{n}^2+1}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2+1}\)
3 \(\mathrm{m}^2=\frac{\mathrm{n}^2+1}{\mathrm{n}^2}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2+1}{\mathrm{~m}^2}\)
4 \(\mathrm{m}^2=\frac{\mathrm{n}^2-1}{\mathrm{n}^2}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2-1}{\mathrm{~m}^2}\)
Explanation:
A Given that,
\(\frac{\mathrm{x}}{\mathrm{ma}}+\frac{\mathrm{y}}{\mathrm{nb}}=1\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1,\)
The line is \(y=-\frac{n b}{m a} x-n b\).
It will touch the ellipse if,
\((-n b)^2=a^2\left(-\frac{n b}{m a}\right)^2+b^2 {\left[c^2=a^2 m^2+b^2\right]}\)
\(n^2=\frac{n^2}{m^2}+1\)
\(m^2=\frac{n^2}{n^2-1} \text { or } \mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2-1}\)
BITSAT-2020
Ellipse
120498
If the area of the ellipse is \(\frac{x^2}{25}+\frac{y^2}{\lambda^2}=1\) is \(20 \pi\) square units, then \(\lambda\) is
1 \(\pm 4\)
2 \(\pm 3\)
3 \(\pm 2\).
4 \(\pm 1\)
Explanation:
(B) Given that,
Equation of focal chord, \(x=2 y+3\),
Eccentricity \(=\frac{3}{4}\)
Foci \(( \pm \mathrm{ae}, 0)\)
Since, \(x=2 y+3\) is a focal chord.
\(\therefore\) It passes through foci.
ae \(=2(0)+3\)
\(\mathrm{a}=\frac{3}{\mathrm{e}}=\frac{3 \times 4}{3}=4\)
Now, \(b^2=a^2\left(1-e^2\right)\)
\(\mathrm{b}^2=(4)^2\left(1-\left(\frac{3}{4}\right)^2\right)=7\)
\(\mathrm{b}= \pm \sqrt{7}\)
\(\therefore\) Length of major axis, \(2 \mathrm{a}=8\)
And length of minor axis, \(2 b=2 \sqrt{7}\)
120495
The length of the latus rectum of \(3 x^2-4 y+6 x-3=0\) is
1 \(\frac{3}{4}\)
2 \(\frac{4}{3}\)
3 2
4 3
Explanation:
B Given, length of the latus rectum of
\(3 x^2-4 y+6 x-3=0\)
\(3 x^2+6 x-3=4 y\)
\(3\left(x^2+2 x-1\right)=4 y\)
\(3\left(x^2+2 x+1-2\right)=4 y\)
\(3(x+1)^2-6=4 y\)
\(3(x+1)^2=4 y+6\)
\((x+1)^2=\frac{4}{3}(y+3 / 2)\)
Let, \(\quad x^2=\frac{4}{3} y\)
Where, \(\mathrm{X}=\mathrm{x}+1\) and \(\mathrm{Y}=\mathrm{y}+\frac{3}{2}\)
And \(\quad 4 \mathrm{~b}=\frac{4}{3}\)
\(\mathrm{b}=\frac{1}{3}\)So, the length of latusrectum is \(\frac{4}{3}\)
Karnataka CET-2011
Ellipse
120496
Eccentricity of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) if it passes through point \((9,5)\) and \((12,4)\) is
1 \(\sqrt{3 / 4}\)
2 \(\sqrt{4 / 5}\)
3 \(\sqrt{5 / 6}\)
4 \(\sqrt{6 / 7}\)
Explanation:
D We have \(\frac{81}{\mathrm{a}^2}+\frac{25}{\mathrm{~b}^2}=1\)
\(\frac{144}{\mathrm{a}^2}+\frac{16}{\mathrm{~b}^2}=1\)
From equation (ii) - equation (i),
\(\frac{63}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=0 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{7}\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1-\frac{1}{7}}=\sqrt{\frac{6}{7}}\)
BITSAT-2018
Ellipse
120497
If \(\frac{x}{m a}+\frac{y}{n b}=1\) touches the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), then
1 \(\mathrm{m}^2=\frac{\mathrm{n}^2}{\mathrm{n}^2-1}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2-1}\)
2 \(\mathrm{m}^2=\frac{\mathrm{n}^2}{\mathrm{n}^2+1}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2+1}\)
3 \(\mathrm{m}^2=\frac{\mathrm{n}^2+1}{\mathrm{n}^2}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2+1}{\mathrm{~m}^2}\)
4 \(\mathrm{m}^2=\frac{\mathrm{n}^2-1}{\mathrm{n}^2}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2-1}{\mathrm{~m}^2}\)
Explanation:
A Given that,
\(\frac{\mathrm{x}}{\mathrm{ma}}+\frac{\mathrm{y}}{\mathrm{nb}}=1\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1,\)
The line is \(y=-\frac{n b}{m a} x-n b\).
It will touch the ellipse if,
\((-n b)^2=a^2\left(-\frac{n b}{m a}\right)^2+b^2 {\left[c^2=a^2 m^2+b^2\right]}\)
\(n^2=\frac{n^2}{m^2}+1\)
\(m^2=\frac{n^2}{n^2-1} \text { or } \mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2-1}\)
BITSAT-2020
Ellipse
120498
If the area of the ellipse is \(\frac{x^2}{25}+\frac{y^2}{\lambda^2}=1\) is \(20 \pi\) square units, then \(\lambda\) is
1 \(\pm 4\)
2 \(\pm 3\)
3 \(\pm 2\).
4 \(\pm 1\)
Explanation:
(B) Given that,
Equation of focal chord, \(x=2 y+3\),
Eccentricity \(=\frac{3}{4}\)
Foci \(( \pm \mathrm{ae}, 0)\)
Since, \(x=2 y+3\) is a focal chord.
\(\therefore\) It passes through foci.
ae \(=2(0)+3\)
\(\mathrm{a}=\frac{3}{\mathrm{e}}=\frac{3 \times 4}{3}=4\)
Now, \(b^2=a^2\left(1-e^2\right)\)
\(\mathrm{b}^2=(4)^2\left(1-\left(\frac{3}{4}\right)^2\right)=7\)
\(\mathrm{b}= \pm \sqrt{7}\)
\(\therefore\) Length of major axis, \(2 \mathrm{a}=8\)
And length of minor axis, \(2 b=2 \sqrt{7}\)
120495
The length of the latus rectum of \(3 x^2-4 y+6 x-3=0\) is
1 \(\frac{3}{4}\)
2 \(\frac{4}{3}\)
3 2
4 3
Explanation:
B Given, length of the latus rectum of
\(3 x^2-4 y+6 x-3=0\)
\(3 x^2+6 x-3=4 y\)
\(3\left(x^2+2 x-1\right)=4 y\)
\(3\left(x^2+2 x+1-2\right)=4 y\)
\(3(x+1)^2-6=4 y\)
\(3(x+1)^2=4 y+6\)
\((x+1)^2=\frac{4}{3}(y+3 / 2)\)
Let, \(\quad x^2=\frac{4}{3} y\)
Where, \(\mathrm{X}=\mathrm{x}+1\) and \(\mathrm{Y}=\mathrm{y}+\frac{3}{2}\)
And \(\quad 4 \mathrm{~b}=\frac{4}{3}\)
\(\mathrm{b}=\frac{1}{3}\)So, the length of latusrectum is \(\frac{4}{3}\)
Karnataka CET-2011
Ellipse
120496
Eccentricity of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) if it passes through point \((9,5)\) and \((12,4)\) is
1 \(\sqrt{3 / 4}\)
2 \(\sqrt{4 / 5}\)
3 \(\sqrt{5 / 6}\)
4 \(\sqrt{6 / 7}\)
Explanation:
D We have \(\frac{81}{\mathrm{a}^2}+\frac{25}{\mathrm{~b}^2}=1\)
\(\frac{144}{\mathrm{a}^2}+\frac{16}{\mathrm{~b}^2}=1\)
From equation (ii) - equation (i),
\(\frac{63}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=0 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{7}\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1-\frac{1}{7}}=\sqrt{\frac{6}{7}}\)
BITSAT-2018
Ellipse
120497
If \(\frac{x}{m a}+\frac{y}{n b}=1\) touches the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), then
1 \(\mathrm{m}^2=\frac{\mathrm{n}^2}{\mathrm{n}^2-1}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2-1}\)
2 \(\mathrm{m}^2=\frac{\mathrm{n}^2}{\mathrm{n}^2+1}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2+1}\)
3 \(\mathrm{m}^2=\frac{\mathrm{n}^2+1}{\mathrm{n}^2}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2+1}{\mathrm{~m}^2}\)
4 \(\mathrm{m}^2=\frac{\mathrm{n}^2-1}{\mathrm{n}^2}\) or \(\mathrm{n}^2=\frac{\mathrm{m}^2-1}{\mathrm{~m}^2}\)
Explanation:
A Given that,
\(\frac{\mathrm{x}}{\mathrm{ma}}+\frac{\mathrm{y}}{\mathrm{nb}}=1\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1,\)
The line is \(y=-\frac{n b}{m a} x-n b\).
It will touch the ellipse if,
\((-n b)^2=a^2\left(-\frac{n b}{m a}\right)^2+b^2 {\left[c^2=a^2 m^2+b^2\right]}\)
\(n^2=\frac{n^2}{m^2}+1\)
\(m^2=\frac{n^2}{n^2-1} \text { or } \mathrm{n}^2=\frac{\mathrm{m}^2}{\mathrm{~m}^2-1}\)
BITSAT-2020
Ellipse
120498
If the area of the ellipse is \(\frac{x^2}{25}+\frac{y^2}{\lambda^2}=1\) is \(20 \pi\) square units, then \(\lambda\) is
1 \(\pm 4\)
2 \(\pm 3\)
3 \(\pm 2\).
4 \(\pm 1\)
Explanation:
(B) Given that,
Equation of focal chord, \(x=2 y+3\),
Eccentricity \(=\frac{3}{4}\)
Foci \(( \pm \mathrm{ae}, 0)\)
Since, \(x=2 y+3\) is a focal chord.
\(\therefore\) It passes through foci.
ae \(=2(0)+3\)
\(\mathrm{a}=\frac{3}{\mathrm{e}}=\frac{3 \times 4}{3}=4\)
Now, \(b^2=a^2\left(1-e^2\right)\)
\(\mathrm{b}^2=(4)^2\left(1-\left(\frac{3}{4}\right)^2\right)=7\)
\(\mathrm{b}= \pm \sqrt{7}\)
\(\therefore\) Length of major axis, \(2 \mathrm{a}=8\)
And length of minor axis, \(2 b=2 \sqrt{7}\)
