A Given,
\(2 y^2+5 x-6 y+1=0\)
\(y^2-3 y+\frac{5}{2} x+\frac{1}{2}=0\)
\(y^2-3 y+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+\frac{5}{2} x+\frac{1}{2}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2} x-\frac{9}{4}+\frac{1}{2}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2} x-\frac{7}{4}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2}\left(x-\frac{7}{10}\right)=0\)
\(\left(y-\frac{3}{2}\right)^2=-\frac{5}{2}\left(x-\frac{7}{10}\right)\)
\(\left(y-\frac{3}{2}\right)^2=-4 \times \frac{5}{8}\left(x-\frac{7}{10}\right)\)
Hence, vertex is \(\left(\frac{7}{10}, \frac{3}{2}\right)\)
and focus is \(\left(\frac{3}{40}, \frac{3}{2}\right)\)
AP EAMCET-20.04.2019
Parabola
120087
The parabola \(x^2=4\) ay makes an intercept of length \(\sqrt{40}\) unit on the line \(y=1+2 x\), then a value of \(a\) is
1 2
2 -2
3 -1
4 4
Explanation:
B Given,
\(x^2=4 a y\) and \(y=1+2 x\)
On solving equation (i) and (ii), we get -
\(x^2=4 a(1+2 x)\)
\(x^2=4 a+8 a x\)
\(x^2-8 a x-4 a=0\)
So, the given line cuts the parabola at two points \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\)
\(x_1+x_2=8 a \text { and } x_1, x_2=-4 a\)
\((\sqrt{40})^2=\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2\)
\(40=\left(x_1-x_2\right)^2+\left(\frac{x_1^2}{4 a}-\frac{x_2^2}{4 a}\right)^2\)
\(40=\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right]\left[1+\frac{\left(x_1+x_2\right)^2}{16 a^2}\right]\)
\(40=(8 a)^2-4(-4 a)\left[1+\frac{(8 a)^2}{16 a^2}\right]\)
\(40=8 a^2+16 a\left[1+\frac{64 a^2}{16 a^2}\right]\)
\(40=5\left(8 a^2+16 a\right)\)
\(40=40 a^2+80 a\)
\(1=a^2+2 a\)
\(a^2+2 a-1=0\)
\((a-1)(a+2)=0\)
\(a=1,-2\)
AP EAMCET-20.04.2019
Parabola
120088
The co-ordinates of focus of the parabola \(5 x^2=\) \(-12 \mathrm{y}\) are
1 \(\left(\frac{3}{5}, 0\right)\)
2 \(\left(\frac{-3}{5}, 0\right)\)
3 \(\left(0, \frac{3}{5}\right)\)
4 \(\left(0, \frac{-3}{5}\right)\)
Explanation:
D Given,
\(5 x^2=-12 y\)
or
\((x-0)^2=\frac{-12}{5} y\)
or
\((x-0)^2=4\left(\frac{-3}{5}\right) y\)Hence, focus is - \(\left(0, \frac{-3}{5}\right)\)
AP EAMCET-18.09.2020
Parabola
120089
If a chord of the parabola \(y^2=4 x\) passes through its focus and makes an angle \(\theta\) with the \(\mathrm{X}\)-axis, then its length is
A Given,
\(2 y^2+5 x-6 y+1=0\)
\(y^2-3 y+\frac{5}{2} x+\frac{1}{2}=0\)
\(y^2-3 y+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+\frac{5}{2} x+\frac{1}{2}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2} x-\frac{9}{4}+\frac{1}{2}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2} x-\frac{7}{4}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2}\left(x-\frac{7}{10}\right)=0\)
\(\left(y-\frac{3}{2}\right)^2=-\frac{5}{2}\left(x-\frac{7}{10}\right)\)
\(\left(y-\frac{3}{2}\right)^2=-4 \times \frac{5}{8}\left(x-\frac{7}{10}\right)\)
Hence, vertex is \(\left(\frac{7}{10}, \frac{3}{2}\right)\)
and focus is \(\left(\frac{3}{40}, \frac{3}{2}\right)\)
AP EAMCET-20.04.2019
Parabola
120087
The parabola \(x^2=4\) ay makes an intercept of length \(\sqrt{40}\) unit on the line \(y=1+2 x\), then a value of \(a\) is
1 2
2 -2
3 -1
4 4
Explanation:
B Given,
\(x^2=4 a y\) and \(y=1+2 x\)
On solving equation (i) and (ii), we get -
\(x^2=4 a(1+2 x)\)
\(x^2=4 a+8 a x\)
\(x^2-8 a x-4 a=0\)
So, the given line cuts the parabola at two points \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\)
\(x_1+x_2=8 a \text { and } x_1, x_2=-4 a\)
\((\sqrt{40})^2=\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2\)
\(40=\left(x_1-x_2\right)^2+\left(\frac{x_1^2}{4 a}-\frac{x_2^2}{4 a}\right)^2\)
\(40=\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right]\left[1+\frac{\left(x_1+x_2\right)^2}{16 a^2}\right]\)
\(40=(8 a)^2-4(-4 a)\left[1+\frac{(8 a)^2}{16 a^2}\right]\)
\(40=8 a^2+16 a\left[1+\frac{64 a^2}{16 a^2}\right]\)
\(40=5\left(8 a^2+16 a\right)\)
\(40=40 a^2+80 a\)
\(1=a^2+2 a\)
\(a^2+2 a-1=0\)
\((a-1)(a+2)=0\)
\(a=1,-2\)
AP EAMCET-20.04.2019
Parabola
120088
The co-ordinates of focus of the parabola \(5 x^2=\) \(-12 \mathrm{y}\) are
1 \(\left(\frac{3}{5}, 0\right)\)
2 \(\left(\frac{-3}{5}, 0\right)\)
3 \(\left(0, \frac{3}{5}\right)\)
4 \(\left(0, \frac{-3}{5}\right)\)
Explanation:
D Given,
\(5 x^2=-12 y\)
or
\((x-0)^2=\frac{-12}{5} y\)
or
\((x-0)^2=4\left(\frac{-3}{5}\right) y\)Hence, focus is - \(\left(0, \frac{-3}{5}\right)\)
AP EAMCET-18.09.2020
Parabola
120089
If a chord of the parabola \(y^2=4 x\) passes through its focus and makes an angle \(\theta\) with the \(\mathrm{X}\)-axis, then its length is
A Given,
\(2 y^2+5 x-6 y+1=0\)
\(y^2-3 y+\frac{5}{2} x+\frac{1}{2}=0\)
\(y^2-3 y+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+\frac{5}{2} x+\frac{1}{2}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2} x-\frac{9}{4}+\frac{1}{2}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2} x-\frac{7}{4}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2}\left(x-\frac{7}{10}\right)=0\)
\(\left(y-\frac{3}{2}\right)^2=-\frac{5}{2}\left(x-\frac{7}{10}\right)\)
\(\left(y-\frac{3}{2}\right)^2=-4 \times \frac{5}{8}\left(x-\frac{7}{10}\right)\)
Hence, vertex is \(\left(\frac{7}{10}, \frac{3}{2}\right)\)
and focus is \(\left(\frac{3}{40}, \frac{3}{2}\right)\)
AP EAMCET-20.04.2019
Parabola
120087
The parabola \(x^2=4\) ay makes an intercept of length \(\sqrt{40}\) unit on the line \(y=1+2 x\), then a value of \(a\) is
1 2
2 -2
3 -1
4 4
Explanation:
B Given,
\(x^2=4 a y\) and \(y=1+2 x\)
On solving equation (i) and (ii), we get -
\(x^2=4 a(1+2 x)\)
\(x^2=4 a+8 a x\)
\(x^2-8 a x-4 a=0\)
So, the given line cuts the parabola at two points \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\)
\(x_1+x_2=8 a \text { and } x_1, x_2=-4 a\)
\((\sqrt{40})^2=\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2\)
\(40=\left(x_1-x_2\right)^2+\left(\frac{x_1^2}{4 a}-\frac{x_2^2}{4 a}\right)^2\)
\(40=\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right]\left[1+\frac{\left(x_1+x_2\right)^2}{16 a^2}\right]\)
\(40=(8 a)^2-4(-4 a)\left[1+\frac{(8 a)^2}{16 a^2}\right]\)
\(40=8 a^2+16 a\left[1+\frac{64 a^2}{16 a^2}\right]\)
\(40=5\left(8 a^2+16 a\right)\)
\(40=40 a^2+80 a\)
\(1=a^2+2 a\)
\(a^2+2 a-1=0\)
\((a-1)(a+2)=0\)
\(a=1,-2\)
AP EAMCET-20.04.2019
Parabola
120088
The co-ordinates of focus of the parabola \(5 x^2=\) \(-12 \mathrm{y}\) are
1 \(\left(\frac{3}{5}, 0\right)\)
2 \(\left(\frac{-3}{5}, 0\right)\)
3 \(\left(0, \frac{3}{5}\right)\)
4 \(\left(0, \frac{-3}{5}\right)\)
Explanation:
D Given,
\(5 x^2=-12 y\)
or
\((x-0)^2=\frac{-12}{5} y\)
or
\((x-0)^2=4\left(\frac{-3}{5}\right) y\)Hence, focus is - \(\left(0, \frac{-3}{5}\right)\)
AP EAMCET-18.09.2020
Parabola
120089
If a chord of the parabola \(y^2=4 x\) passes through its focus and makes an angle \(\theta\) with the \(\mathrm{X}\)-axis, then its length is
A Given,
\(2 y^2+5 x-6 y+1=0\)
\(y^2-3 y+\frac{5}{2} x+\frac{1}{2}=0\)
\(y^2-3 y+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+\frac{5}{2} x+\frac{1}{2}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2} x-\frac{9}{4}+\frac{1}{2}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2} x-\frac{7}{4}=0\)
\(\left(y-\frac{3}{2}\right)^2+\frac{5}{2}\left(x-\frac{7}{10}\right)=0\)
\(\left(y-\frac{3}{2}\right)^2=-\frac{5}{2}\left(x-\frac{7}{10}\right)\)
\(\left(y-\frac{3}{2}\right)^2=-4 \times \frac{5}{8}\left(x-\frac{7}{10}\right)\)
Hence, vertex is \(\left(\frac{7}{10}, \frac{3}{2}\right)\)
and focus is \(\left(\frac{3}{40}, \frac{3}{2}\right)\)
AP EAMCET-20.04.2019
Parabola
120087
The parabola \(x^2=4\) ay makes an intercept of length \(\sqrt{40}\) unit on the line \(y=1+2 x\), then a value of \(a\) is
1 2
2 -2
3 -1
4 4
Explanation:
B Given,
\(x^2=4 a y\) and \(y=1+2 x\)
On solving equation (i) and (ii), we get -
\(x^2=4 a(1+2 x)\)
\(x^2=4 a+8 a x\)
\(x^2-8 a x-4 a=0\)
So, the given line cuts the parabola at two points \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\)
\(x_1+x_2=8 a \text { and } x_1, x_2=-4 a\)
\((\sqrt{40})^2=\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2\)
\(40=\left(x_1-x_2\right)^2+\left(\frac{x_1^2}{4 a}-\frac{x_2^2}{4 a}\right)^2\)
\(40=\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right]\left[1+\frac{\left(x_1+x_2\right)^2}{16 a^2}\right]\)
\(40=(8 a)^2-4(-4 a)\left[1+\frac{(8 a)^2}{16 a^2}\right]\)
\(40=8 a^2+16 a\left[1+\frac{64 a^2}{16 a^2}\right]\)
\(40=5\left(8 a^2+16 a\right)\)
\(40=40 a^2+80 a\)
\(1=a^2+2 a\)
\(a^2+2 a-1=0\)
\((a-1)(a+2)=0\)
\(a=1,-2\)
AP EAMCET-20.04.2019
Parabola
120088
The co-ordinates of focus of the parabola \(5 x^2=\) \(-12 \mathrm{y}\) are
1 \(\left(\frac{3}{5}, 0\right)\)
2 \(\left(\frac{-3}{5}, 0\right)\)
3 \(\left(0, \frac{3}{5}\right)\)
4 \(\left(0, \frac{-3}{5}\right)\)
Explanation:
D Given,
\(5 x^2=-12 y\)
or
\((x-0)^2=\frac{-12}{5} y\)
or
\((x-0)^2=4\left(\frac{-3}{5}\right) y\)Hence, focus is - \(\left(0, \frac{-3}{5}\right)\)
AP EAMCET-18.09.2020
Parabola
120089
If a chord of the parabola \(y^2=4 x\) passes through its focus and makes an angle \(\theta\) with the \(\mathrm{X}\)-axis, then its length is
