120836
Equation of a tangent to the hyperbola \(5 x^2-y^2\) \(=5\) and which passes through an external point \((2,8)\) is
1 \(3 \mathrm{x}-\mathrm{y}+2=0\)
2 \(3 \mathrm{x}+\mathrm{y}-14=0\)
3 \(23 \mathrm{x}-3 \mathrm{y}-22=0\)
4 \(3 \mathrm{x}-23 \mathrm{y}+178=0\)
Explanation:
A Given, equation of hyperbola can be write as
\(5 x^2-y^2=5\)
\(\frac{x^2}{1}-\frac{y^2}{(\sqrt{5})^2}=1\)
Here, \(\mathrm{a}=1\) and \(\mathrm{b}=\sqrt{5}\)
Since, \(y=m x \pm \sqrt{a^2 m^2-b^2}\)
\(y=m x \pm \sqrt{m^2-5}\)
\((8-2 m)^2=\left(\sqrt{m^2-5}\right)^2\)
\((8-2 m)^2=m^2-5\)
\(64+4 m^2-32 m=m^2-5\)
\(3 m^2-32 m+69=0\)
\(m=\frac{23}{3}, 3\)
Hence, the equation of tangent are -
\(3 x-y+2=0 \text { and } 23 y-3 y-22=0\)
WB JEE-2019
Hyperbola
120837
Consider the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) the area of the triangle formed by the asymptotes and the tangent drawn to it at \((a, 0)\) is
1 \(\frac{1}{2} \mathrm{ab}\)
2 ab
3 \(2 \mathrm{ab}\)
4 \(4 \mathrm{ab}\)
Explanation:
B Equation of tangent at \((a, 0)\) is \(x=a\) and the point of intersection of \(x=a \&\) the asymptotes will be obtained by solving \(\mathrm{x}=0\) \& equation of asymptotes
\(b x \pm a y=0\)
\(\therefore a b \pm a y=0 \Rightarrow y= \pm b\)
Now, point of intersection are \((0,0),(a, b)\) and \((a,-b)\)
\(\therefore\) Area of triangle \(=\frac{1}{2} \times(2 b \times a)=a b\) sq. units.
Manipal UGET-2017
Hyperbola
120838
Asymptotes of a Hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) are
1 \(x= \pm \frac{4}{5} y\)
2 \(y= \pm \frac{4}{5} x\)
3 \(x= \pm \frac{25}{16} y\)
4 \(y= \pm \frac{5}{4} x\)
Explanation:
B Given, \(\frac{\mathrm{x}^2}{25}-\frac{\mathrm{y}^2}{16}=1\)
On comparing with \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
Center \(\equiv(\mathrm{h}, \mathrm{k})=(0,0)\)
\(a^2=25 \Rightarrow a=5\)
and
\(b^2=16 \Rightarrow b=4\)
Now,
Asymptotes of hyperbola are -
\(y= \pm \frac{b(x-h)}{a}+k\)
\(y= \pm \frac{4(x-0)}{5}+0\)
\(y= \pm \frac{4}{5} x\)
GUJCET-2009
Hyperbola
120839
Find the measure of angle between the asymptotes of \(x^2-y^2=16\).
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{2}\)
Explanation:
D : Given hyperbola,
\(x^2-y^2=16\)
\(\frac{x^2}{16}-\frac{y^2}{16}=1\)
Here, \(\quad \mathrm{a}=4, \quad \mathrm{~b}=4\)
Angle between asymptotes of hyperbola -
\(\frac{x^2}{16}-\frac{y^2}{16}=1 \text { is } 2 \tan ^{-1}\left(\frac{b}{a}\right)\)
So, Angle \(=2 \tan ^{-1}\left(\frac{4}{4}\right)\)
\(=2 \tan ^{-1}(1)\)
\(=2 \tan ^{-1}\left(\tan \frac{\pi}{4}\right)=2 \times \frac{\pi}{4}=\frac{\pi}{2}\)
120836
Equation of a tangent to the hyperbola \(5 x^2-y^2\) \(=5\) and which passes through an external point \((2,8)\) is
1 \(3 \mathrm{x}-\mathrm{y}+2=0\)
2 \(3 \mathrm{x}+\mathrm{y}-14=0\)
3 \(23 \mathrm{x}-3 \mathrm{y}-22=0\)
4 \(3 \mathrm{x}-23 \mathrm{y}+178=0\)
Explanation:
A Given, equation of hyperbola can be write as
\(5 x^2-y^2=5\)
\(\frac{x^2}{1}-\frac{y^2}{(\sqrt{5})^2}=1\)
Here, \(\mathrm{a}=1\) and \(\mathrm{b}=\sqrt{5}\)
Since, \(y=m x \pm \sqrt{a^2 m^2-b^2}\)
\(y=m x \pm \sqrt{m^2-5}\)
\((8-2 m)^2=\left(\sqrt{m^2-5}\right)^2\)
\((8-2 m)^2=m^2-5\)
\(64+4 m^2-32 m=m^2-5\)
\(3 m^2-32 m+69=0\)
\(m=\frac{23}{3}, 3\)
Hence, the equation of tangent are -
\(3 x-y+2=0 \text { and } 23 y-3 y-22=0\)
WB JEE-2019
Hyperbola
120837
Consider the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) the area of the triangle formed by the asymptotes and the tangent drawn to it at \((a, 0)\) is
1 \(\frac{1}{2} \mathrm{ab}\)
2 ab
3 \(2 \mathrm{ab}\)
4 \(4 \mathrm{ab}\)
Explanation:
B Equation of tangent at \((a, 0)\) is \(x=a\) and the point of intersection of \(x=a \&\) the asymptotes will be obtained by solving \(\mathrm{x}=0\) \& equation of asymptotes
\(b x \pm a y=0\)
\(\therefore a b \pm a y=0 \Rightarrow y= \pm b\)
Now, point of intersection are \((0,0),(a, b)\) and \((a,-b)\)
\(\therefore\) Area of triangle \(=\frac{1}{2} \times(2 b \times a)=a b\) sq. units.
Manipal UGET-2017
Hyperbola
120838
Asymptotes of a Hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) are
1 \(x= \pm \frac{4}{5} y\)
2 \(y= \pm \frac{4}{5} x\)
3 \(x= \pm \frac{25}{16} y\)
4 \(y= \pm \frac{5}{4} x\)
Explanation:
B Given, \(\frac{\mathrm{x}^2}{25}-\frac{\mathrm{y}^2}{16}=1\)
On comparing with \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
Center \(\equiv(\mathrm{h}, \mathrm{k})=(0,0)\)
\(a^2=25 \Rightarrow a=5\)
and
\(b^2=16 \Rightarrow b=4\)
Now,
Asymptotes of hyperbola are -
\(y= \pm \frac{b(x-h)}{a}+k\)
\(y= \pm \frac{4(x-0)}{5}+0\)
\(y= \pm \frac{4}{5} x\)
GUJCET-2009
Hyperbola
120839
Find the measure of angle between the asymptotes of \(x^2-y^2=16\).
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{2}\)
Explanation:
D : Given hyperbola,
\(x^2-y^2=16\)
\(\frac{x^2}{16}-\frac{y^2}{16}=1\)
Here, \(\quad \mathrm{a}=4, \quad \mathrm{~b}=4\)
Angle between asymptotes of hyperbola -
\(\frac{x^2}{16}-\frac{y^2}{16}=1 \text { is } 2 \tan ^{-1}\left(\frac{b}{a}\right)\)
So, Angle \(=2 \tan ^{-1}\left(\frac{4}{4}\right)\)
\(=2 \tan ^{-1}(1)\)
\(=2 \tan ^{-1}\left(\tan \frac{\pi}{4}\right)=2 \times \frac{\pi}{4}=\frac{\pi}{2}\)
120836
Equation of a tangent to the hyperbola \(5 x^2-y^2\) \(=5\) and which passes through an external point \((2,8)\) is
1 \(3 \mathrm{x}-\mathrm{y}+2=0\)
2 \(3 \mathrm{x}+\mathrm{y}-14=0\)
3 \(23 \mathrm{x}-3 \mathrm{y}-22=0\)
4 \(3 \mathrm{x}-23 \mathrm{y}+178=0\)
Explanation:
A Given, equation of hyperbola can be write as
\(5 x^2-y^2=5\)
\(\frac{x^2}{1}-\frac{y^2}{(\sqrt{5})^2}=1\)
Here, \(\mathrm{a}=1\) and \(\mathrm{b}=\sqrt{5}\)
Since, \(y=m x \pm \sqrt{a^2 m^2-b^2}\)
\(y=m x \pm \sqrt{m^2-5}\)
\((8-2 m)^2=\left(\sqrt{m^2-5}\right)^2\)
\((8-2 m)^2=m^2-5\)
\(64+4 m^2-32 m=m^2-5\)
\(3 m^2-32 m+69=0\)
\(m=\frac{23}{3}, 3\)
Hence, the equation of tangent are -
\(3 x-y+2=0 \text { and } 23 y-3 y-22=0\)
WB JEE-2019
Hyperbola
120837
Consider the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) the area of the triangle formed by the asymptotes and the tangent drawn to it at \((a, 0)\) is
1 \(\frac{1}{2} \mathrm{ab}\)
2 ab
3 \(2 \mathrm{ab}\)
4 \(4 \mathrm{ab}\)
Explanation:
B Equation of tangent at \((a, 0)\) is \(x=a\) and the point of intersection of \(x=a \&\) the asymptotes will be obtained by solving \(\mathrm{x}=0\) \& equation of asymptotes
\(b x \pm a y=0\)
\(\therefore a b \pm a y=0 \Rightarrow y= \pm b\)
Now, point of intersection are \((0,0),(a, b)\) and \((a,-b)\)
\(\therefore\) Area of triangle \(=\frac{1}{2} \times(2 b \times a)=a b\) sq. units.
Manipal UGET-2017
Hyperbola
120838
Asymptotes of a Hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) are
1 \(x= \pm \frac{4}{5} y\)
2 \(y= \pm \frac{4}{5} x\)
3 \(x= \pm \frac{25}{16} y\)
4 \(y= \pm \frac{5}{4} x\)
Explanation:
B Given, \(\frac{\mathrm{x}^2}{25}-\frac{\mathrm{y}^2}{16}=1\)
On comparing with \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
Center \(\equiv(\mathrm{h}, \mathrm{k})=(0,0)\)
\(a^2=25 \Rightarrow a=5\)
and
\(b^2=16 \Rightarrow b=4\)
Now,
Asymptotes of hyperbola are -
\(y= \pm \frac{b(x-h)}{a}+k\)
\(y= \pm \frac{4(x-0)}{5}+0\)
\(y= \pm \frac{4}{5} x\)
GUJCET-2009
Hyperbola
120839
Find the measure of angle between the asymptotes of \(x^2-y^2=16\).
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{2}\)
Explanation:
D : Given hyperbola,
\(x^2-y^2=16\)
\(\frac{x^2}{16}-\frac{y^2}{16}=1\)
Here, \(\quad \mathrm{a}=4, \quad \mathrm{~b}=4\)
Angle between asymptotes of hyperbola -
\(\frac{x^2}{16}-\frac{y^2}{16}=1 \text { is } 2 \tan ^{-1}\left(\frac{b}{a}\right)\)
So, Angle \(=2 \tan ^{-1}\left(\frac{4}{4}\right)\)
\(=2 \tan ^{-1}(1)\)
\(=2 \tan ^{-1}\left(\tan \frac{\pi}{4}\right)=2 \times \frac{\pi}{4}=\frac{\pi}{2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Hyperbola
120836
Equation of a tangent to the hyperbola \(5 x^2-y^2\) \(=5\) and which passes through an external point \((2,8)\) is
1 \(3 \mathrm{x}-\mathrm{y}+2=0\)
2 \(3 \mathrm{x}+\mathrm{y}-14=0\)
3 \(23 \mathrm{x}-3 \mathrm{y}-22=0\)
4 \(3 \mathrm{x}-23 \mathrm{y}+178=0\)
Explanation:
A Given, equation of hyperbola can be write as
\(5 x^2-y^2=5\)
\(\frac{x^2}{1}-\frac{y^2}{(\sqrt{5})^2}=1\)
Here, \(\mathrm{a}=1\) and \(\mathrm{b}=\sqrt{5}\)
Since, \(y=m x \pm \sqrt{a^2 m^2-b^2}\)
\(y=m x \pm \sqrt{m^2-5}\)
\((8-2 m)^2=\left(\sqrt{m^2-5}\right)^2\)
\((8-2 m)^2=m^2-5\)
\(64+4 m^2-32 m=m^2-5\)
\(3 m^2-32 m+69=0\)
\(m=\frac{23}{3}, 3\)
Hence, the equation of tangent are -
\(3 x-y+2=0 \text { and } 23 y-3 y-22=0\)
WB JEE-2019
Hyperbola
120837
Consider the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) the area of the triangle formed by the asymptotes and the tangent drawn to it at \((a, 0)\) is
1 \(\frac{1}{2} \mathrm{ab}\)
2 ab
3 \(2 \mathrm{ab}\)
4 \(4 \mathrm{ab}\)
Explanation:
B Equation of tangent at \((a, 0)\) is \(x=a\) and the point of intersection of \(x=a \&\) the asymptotes will be obtained by solving \(\mathrm{x}=0\) \& equation of asymptotes
\(b x \pm a y=0\)
\(\therefore a b \pm a y=0 \Rightarrow y= \pm b\)
Now, point of intersection are \((0,0),(a, b)\) and \((a,-b)\)
\(\therefore\) Area of triangle \(=\frac{1}{2} \times(2 b \times a)=a b\) sq. units.
Manipal UGET-2017
Hyperbola
120838
Asymptotes of a Hyperbola \(\frac{x^2}{25}-\frac{y^2}{16}=1\) are
1 \(x= \pm \frac{4}{5} y\)
2 \(y= \pm \frac{4}{5} x\)
3 \(x= \pm \frac{25}{16} y\)
4 \(y= \pm \frac{5}{4} x\)
Explanation:
B Given, \(\frac{\mathrm{x}^2}{25}-\frac{\mathrm{y}^2}{16}=1\)
On comparing with \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
Center \(\equiv(\mathrm{h}, \mathrm{k})=(0,0)\)
\(a^2=25 \Rightarrow a=5\)
and
\(b^2=16 \Rightarrow b=4\)
Now,
Asymptotes of hyperbola are -
\(y= \pm \frac{b(x-h)}{a}+k\)
\(y= \pm \frac{4(x-0)}{5}+0\)
\(y= \pm \frac{4}{5} x\)
GUJCET-2009
Hyperbola
120839
Find the measure of angle between the asymptotes of \(x^2-y^2=16\).
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{2}\)
Explanation:
D : Given hyperbola,
\(x^2-y^2=16\)
\(\frac{x^2}{16}-\frac{y^2}{16}=1\)
Here, \(\quad \mathrm{a}=4, \quad \mathrm{~b}=4\)
Angle between asymptotes of hyperbola -
\(\frac{x^2}{16}-\frac{y^2}{16}=1 \text { is } 2 \tan ^{-1}\left(\frac{b}{a}\right)\)
So, Angle \(=2 \tan ^{-1}\left(\frac{4}{4}\right)\)
\(=2 \tan ^{-1}(1)\)
\(=2 \tan ^{-1}\left(\tan \frac{\pi}{4}\right)=2 \times \frac{\pi}{4}=\frac{\pi}{2}\)
