NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120820
The locus of the point of intersection on the line \(\sqrt{3 x}-y-4 \sqrt{3 k}=0\) and \(\sqrt{3} k x+k y-4 \sqrt{3}=0\) for different real values of \(k\) is a hyperbola \(H\). If \(e\) is the eccentricity of \(\mathrm{H}\) then \(4 \mathrm{e}^2=\)
1 48
2 39
3 13
4 16
Explanation:
D Given line \(\sqrt{3} \mathrm{x}-\mathrm{y}-4 \sqrt{3} \mathrm{~K}=0\)
\(\sqrt{3} \mathrm{x}-\mathrm{y}=4 \sqrt{3} \mathrm{k}\)
\(\mathrm{k}=\frac{\sqrt{3} \mathrm{x}-\mathrm{y}}{4 \sqrt{3}}\)
\(\sqrt{3} \mathrm{kx}+\mathrm{ky}-4 \sqrt{3}=0\)
\(\mathrm{k}(\sqrt{3} \mathrm{x}+\mathrm{y})-4 \sqrt{3}=0\)
\(\mathrm{k}=\frac{4 \sqrt{3}}{\sqrt{3} \mathrm{x}+\mathrm{y}}\)
From equation (i) and (ii) the locus of point of intersection of the line
\(\frac{\sqrt{3} x-y}{4 \sqrt{3}}=\frac{4 \sqrt{3}}{\sqrt{3} x+y}\)
\((\sqrt{3} x)^2-y^2=16 \times 3\)
\(3 x^2-y^2=48\)
\(\frac{x^2}{16}-\frac{y^2}{48}=1\)
Then eccentricity of hyperbola,
\(b^2=a^2\left(e^2-1\right)\)
Comparing, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\mathrm{a}^2=16, \mathrm{~b}^2=48\)
\(48=16\left(\mathrm{e}^2-1\right)\)
\(3=\mathrm{e}^2-1\)
\(\mathrm{e}^2=4\)
Now,
\(=4 \mathrm{e}^2\)
\(=4 \times 4\)
\(=16\)
AP EAMCET-07.07.2022
Hyperbola
120821
Find the equation of tangent to the hyperbola \(4 x^2-y^2=64\), which is parallel to the line \(8 \mathrm{x}-6 \mathrm{y}+\mathbf{1 1}=\mathbf{0}\)
1 \(2 x+y=1\)
2 Tangent does not exists
3 \(3 x+y=1\)
4 \(x+3 y=1\)
Explanation:
B \(4 \mathrm{x}^2-\mathrm{y}^2=64\)
\(\frac{x^2}{16}-\frac{y^2}{64}=1\)
Equation of tangent in slope form
\(\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2}\)
\(8 \mathrm{x}-6 \mathrm{y}+11=0\)
\(8 \mathrm{x}-6 \mathrm{y}=-11\)
\(8 x-6 y=-11\)
\(\text { Slope }(m)=-\frac{x}{y}\)
\(=-\frac{8}{(-6)}=\frac{4}{3}\)
\((\mathrm{~m})=\frac{4}{3}\)
Line
\(8 x-6 y=-11\)
\(\mathrm{a}^2=16 \quad \mathrm{~b}^2=64\)
\(y=\frac{4}{3} x \pm \sqrt{16 \times \frac{16}{9}-64}\)
\(y=\frac{4}{3} x \pm \sqrt{\frac{256-576}{9}}\)
\(y=\frac{4}{3} x \pm \sqrt{\frac{-320}{9}}\), an imaginary value of the constant
term.
So, Tangent does not exist.
GUJCET-2007
Hyperbola
120823
The curve for which the length of the normal is equal to the length of the radius vector, are
1 only circles
2 only rectangular hyperbolas
3 either circles or rectangular hyperbolas
4 None of the above
Explanation:
C We have-
length of normal \(=\) radius vector.
\(\mathrm{y} \sqrt{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2}=\sqrt{\mathrm{x}^2+\mathrm{y}^2}\) \(\mathrm{y}^2\left\{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right\}=\mathrm{x}^2+\mathrm{y}^2\)
\(y^2+y^2\left(\frac{d y}{d x}\right)^2=x^2+y^2\)
\(y^2\left(\frac{d y}{d x}\right)^2=x^2\)
\(x= \pm y \frac{d y}{d x}\)
Here, we have-
\(x=y \frac{d y}{d x} \quad \text { or } \quad x=-y \frac{d y}{d x}\)
Now,
\(\text { If } x=y \frac{d y}{d x}\)
\(x d x-y d y=0\)
\(x^2-y^2=c_1\)
and,
\(x=-y \frac{d y}{d x}\)
\(x d x+y d y=0\)
\(x^2+y^2=c_2\)Clearly, \(x^2-y^2=c_1\) represents a rectangular hyperbola and \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{c}_2\) represents circle.
Manipal UGET-2013
Hyperbola
120822
The line \(x \cos \alpha+y \sin \alpha=p\) touches the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), if:
A The line \(\mathrm{x} \cos \propto+\mathrm{y} \sin \propto=\mathrm{p}\) touches the hyperbola,
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { if }\)
\(a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2\)
120820
The locus of the point of intersection on the line \(\sqrt{3 x}-y-4 \sqrt{3 k}=0\) and \(\sqrt{3} k x+k y-4 \sqrt{3}=0\) for different real values of \(k\) is a hyperbola \(H\). If \(e\) is the eccentricity of \(\mathrm{H}\) then \(4 \mathrm{e}^2=\)
1 48
2 39
3 13
4 16
Explanation:
D Given line \(\sqrt{3} \mathrm{x}-\mathrm{y}-4 \sqrt{3} \mathrm{~K}=0\)
\(\sqrt{3} \mathrm{x}-\mathrm{y}=4 \sqrt{3} \mathrm{k}\)
\(\mathrm{k}=\frac{\sqrt{3} \mathrm{x}-\mathrm{y}}{4 \sqrt{3}}\)
\(\sqrt{3} \mathrm{kx}+\mathrm{ky}-4 \sqrt{3}=0\)
\(\mathrm{k}(\sqrt{3} \mathrm{x}+\mathrm{y})-4 \sqrt{3}=0\)
\(\mathrm{k}=\frac{4 \sqrt{3}}{\sqrt{3} \mathrm{x}+\mathrm{y}}\)
From equation (i) and (ii) the locus of point of intersection of the line
\(\frac{\sqrt{3} x-y}{4 \sqrt{3}}=\frac{4 \sqrt{3}}{\sqrt{3} x+y}\)
\((\sqrt{3} x)^2-y^2=16 \times 3\)
\(3 x^2-y^2=48\)
\(\frac{x^2}{16}-\frac{y^2}{48}=1\)
Then eccentricity of hyperbola,
\(b^2=a^2\left(e^2-1\right)\)
Comparing, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\mathrm{a}^2=16, \mathrm{~b}^2=48\)
\(48=16\left(\mathrm{e}^2-1\right)\)
\(3=\mathrm{e}^2-1\)
\(\mathrm{e}^2=4\)
Now,
\(=4 \mathrm{e}^2\)
\(=4 \times 4\)
\(=16\)
AP EAMCET-07.07.2022
Hyperbola
120821
Find the equation of tangent to the hyperbola \(4 x^2-y^2=64\), which is parallel to the line \(8 \mathrm{x}-6 \mathrm{y}+\mathbf{1 1}=\mathbf{0}\)
1 \(2 x+y=1\)
2 Tangent does not exists
3 \(3 x+y=1\)
4 \(x+3 y=1\)
Explanation:
B \(4 \mathrm{x}^2-\mathrm{y}^2=64\)
\(\frac{x^2}{16}-\frac{y^2}{64}=1\)
Equation of tangent in slope form
\(\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2}\)
\(8 \mathrm{x}-6 \mathrm{y}+11=0\)
\(8 \mathrm{x}-6 \mathrm{y}=-11\)
\(8 x-6 y=-11\)
\(\text { Slope }(m)=-\frac{x}{y}\)
\(=-\frac{8}{(-6)}=\frac{4}{3}\)
\((\mathrm{~m})=\frac{4}{3}\)
Line
\(8 x-6 y=-11\)
\(\mathrm{a}^2=16 \quad \mathrm{~b}^2=64\)
\(y=\frac{4}{3} x \pm \sqrt{16 \times \frac{16}{9}-64}\)
\(y=\frac{4}{3} x \pm \sqrt{\frac{256-576}{9}}\)
\(y=\frac{4}{3} x \pm \sqrt{\frac{-320}{9}}\), an imaginary value of the constant
term.
So, Tangent does not exist.
GUJCET-2007
Hyperbola
120823
The curve for which the length of the normal is equal to the length of the radius vector, are
1 only circles
2 only rectangular hyperbolas
3 either circles or rectangular hyperbolas
4 None of the above
Explanation:
C We have-
length of normal \(=\) radius vector.
\(\mathrm{y} \sqrt{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2}=\sqrt{\mathrm{x}^2+\mathrm{y}^2}\) \(\mathrm{y}^2\left\{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right\}=\mathrm{x}^2+\mathrm{y}^2\)
\(y^2+y^2\left(\frac{d y}{d x}\right)^2=x^2+y^2\)
\(y^2\left(\frac{d y}{d x}\right)^2=x^2\)
\(x= \pm y \frac{d y}{d x}\)
Here, we have-
\(x=y \frac{d y}{d x} \quad \text { or } \quad x=-y \frac{d y}{d x}\)
Now,
\(\text { If } x=y \frac{d y}{d x}\)
\(x d x-y d y=0\)
\(x^2-y^2=c_1\)
and,
\(x=-y \frac{d y}{d x}\)
\(x d x+y d y=0\)
\(x^2+y^2=c_2\)Clearly, \(x^2-y^2=c_1\) represents a rectangular hyperbola and \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{c}_2\) represents circle.
Manipal UGET-2013
Hyperbola
120822
The line \(x \cos \alpha+y \sin \alpha=p\) touches the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), if:
A The line \(\mathrm{x} \cos \propto+\mathrm{y} \sin \propto=\mathrm{p}\) touches the hyperbola,
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { if }\)
\(a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2\)
120820
The locus of the point of intersection on the line \(\sqrt{3 x}-y-4 \sqrt{3 k}=0\) and \(\sqrt{3} k x+k y-4 \sqrt{3}=0\) for different real values of \(k\) is a hyperbola \(H\). If \(e\) is the eccentricity of \(\mathrm{H}\) then \(4 \mathrm{e}^2=\)
1 48
2 39
3 13
4 16
Explanation:
D Given line \(\sqrt{3} \mathrm{x}-\mathrm{y}-4 \sqrt{3} \mathrm{~K}=0\)
\(\sqrt{3} \mathrm{x}-\mathrm{y}=4 \sqrt{3} \mathrm{k}\)
\(\mathrm{k}=\frac{\sqrt{3} \mathrm{x}-\mathrm{y}}{4 \sqrt{3}}\)
\(\sqrt{3} \mathrm{kx}+\mathrm{ky}-4 \sqrt{3}=0\)
\(\mathrm{k}(\sqrt{3} \mathrm{x}+\mathrm{y})-4 \sqrt{3}=0\)
\(\mathrm{k}=\frac{4 \sqrt{3}}{\sqrt{3} \mathrm{x}+\mathrm{y}}\)
From equation (i) and (ii) the locus of point of intersection of the line
\(\frac{\sqrt{3} x-y}{4 \sqrt{3}}=\frac{4 \sqrt{3}}{\sqrt{3} x+y}\)
\((\sqrt{3} x)^2-y^2=16 \times 3\)
\(3 x^2-y^2=48\)
\(\frac{x^2}{16}-\frac{y^2}{48}=1\)
Then eccentricity of hyperbola,
\(b^2=a^2\left(e^2-1\right)\)
Comparing, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\mathrm{a}^2=16, \mathrm{~b}^2=48\)
\(48=16\left(\mathrm{e}^2-1\right)\)
\(3=\mathrm{e}^2-1\)
\(\mathrm{e}^2=4\)
Now,
\(=4 \mathrm{e}^2\)
\(=4 \times 4\)
\(=16\)
AP EAMCET-07.07.2022
Hyperbola
120821
Find the equation of tangent to the hyperbola \(4 x^2-y^2=64\), which is parallel to the line \(8 \mathrm{x}-6 \mathrm{y}+\mathbf{1 1}=\mathbf{0}\)
1 \(2 x+y=1\)
2 Tangent does not exists
3 \(3 x+y=1\)
4 \(x+3 y=1\)
Explanation:
B \(4 \mathrm{x}^2-\mathrm{y}^2=64\)
\(\frac{x^2}{16}-\frac{y^2}{64}=1\)
Equation of tangent in slope form
\(\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2}\)
\(8 \mathrm{x}-6 \mathrm{y}+11=0\)
\(8 \mathrm{x}-6 \mathrm{y}=-11\)
\(8 x-6 y=-11\)
\(\text { Slope }(m)=-\frac{x}{y}\)
\(=-\frac{8}{(-6)}=\frac{4}{3}\)
\((\mathrm{~m})=\frac{4}{3}\)
Line
\(8 x-6 y=-11\)
\(\mathrm{a}^2=16 \quad \mathrm{~b}^2=64\)
\(y=\frac{4}{3} x \pm \sqrt{16 \times \frac{16}{9}-64}\)
\(y=\frac{4}{3} x \pm \sqrt{\frac{256-576}{9}}\)
\(y=\frac{4}{3} x \pm \sqrt{\frac{-320}{9}}\), an imaginary value of the constant
term.
So, Tangent does not exist.
GUJCET-2007
Hyperbola
120823
The curve for which the length of the normal is equal to the length of the radius vector, are
1 only circles
2 only rectangular hyperbolas
3 either circles or rectangular hyperbolas
4 None of the above
Explanation:
C We have-
length of normal \(=\) radius vector.
\(\mathrm{y} \sqrt{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2}=\sqrt{\mathrm{x}^2+\mathrm{y}^2}\) \(\mathrm{y}^2\left\{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right\}=\mathrm{x}^2+\mathrm{y}^2\)
\(y^2+y^2\left(\frac{d y}{d x}\right)^2=x^2+y^2\)
\(y^2\left(\frac{d y}{d x}\right)^2=x^2\)
\(x= \pm y \frac{d y}{d x}\)
Here, we have-
\(x=y \frac{d y}{d x} \quad \text { or } \quad x=-y \frac{d y}{d x}\)
Now,
\(\text { If } x=y \frac{d y}{d x}\)
\(x d x-y d y=0\)
\(x^2-y^2=c_1\)
and,
\(x=-y \frac{d y}{d x}\)
\(x d x+y d y=0\)
\(x^2+y^2=c_2\)Clearly, \(x^2-y^2=c_1\) represents a rectangular hyperbola and \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{c}_2\) represents circle.
Manipal UGET-2013
Hyperbola
120822
The line \(x \cos \alpha+y \sin \alpha=p\) touches the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), if:
A The line \(\mathrm{x} \cos \propto+\mathrm{y} \sin \propto=\mathrm{p}\) touches the hyperbola,
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { if }\)
\(a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2\)
120820
The locus of the point of intersection on the line \(\sqrt{3 x}-y-4 \sqrt{3 k}=0\) and \(\sqrt{3} k x+k y-4 \sqrt{3}=0\) for different real values of \(k\) is a hyperbola \(H\). If \(e\) is the eccentricity of \(\mathrm{H}\) then \(4 \mathrm{e}^2=\)
1 48
2 39
3 13
4 16
Explanation:
D Given line \(\sqrt{3} \mathrm{x}-\mathrm{y}-4 \sqrt{3} \mathrm{~K}=0\)
\(\sqrt{3} \mathrm{x}-\mathrm{y}=4 \sqrt{3} \mathrm{k}\)
\(\mathrm{k}=\frac{\sqrt{3} \mathrm{x}-\mathrm{y}}{4 \sqrt{3}}\)
\(\sqrt{3} \mathrm{kx}+\mathrm{ky}-4 \sqrt{3}=0\)
\(\mathrm{k}(\sqrt{3} \mathrm{x}+\mathrm{y})-4 \sqrt{3}=0\)
\(\mathrm{k}=\frac{4 \sqrt{3}}{\sqrt{3} \mathrm{x}+\mathrm{y}}\)
From equation (i) and (ii) the locus of point of intersection of the line
\(\frac{\sqrt{3} x-y}{4 \sqrt{3}}=\frac{4 \sqrt{3}}{\sqrt{3} x+y}\)
\((\sqrt{3} x)^2-y^2=16 \times 3\)
\(3 x^2-y^2=48\)
\(\frac{x^2}{16}-\frac{y^2}{48}=1\)
Then eccentricity of hyperbola,
\(b^2=a^2\left(e^2-1\right)\)
Comparing, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\mathrm{a}^2=16, \mathrm{~b}^2=48\)
\(48=16\left(\mathrm{e}^2-1\right)\)
\(3=\mathrm{e}^2-1\)
\(\mathrm{e}^2=4\)
Now,
\(=4 \mathrm{e}^2\)
\(=4 \times 4\)
\(=16\)
AP EAMCET-07.07.2022
Hyperbola
120821
Find the equation of tangent to the hyperbola \(4 x^2-y^2=64\), which is parallel to the line \(8 \mathrm{x}-6 \mathrm{y}+\mathbf{1 1}=\mathbf{0}\)
1 \(2 x+y=1\)
2 Tangent does not exists
3 \(3 x+y=1\)
4 \(x+3 y=1\)
Explanation:
B \(4 \mathrm{x}^2-\mathrm{y}^2=64\)
\(\frac{x^2}{16}-\frac{y^2}{64}=1\)
Equation of tangent in slope form
\(\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2}\)
\(8 \mathrm{x}-6 \mathrm{y}+11=0\)
\(8 \mathrm{x}-6 \mathrm{y}=-11\)
\(8 x-6 y=-11\)
\(\text { Slope }(m)=-\frac{x}{y}\)
\(=-\frac{8}{(-6)}=\frac{4}{3}\)
\((\mathrm{~m})=\frac{4}{3}\)
Line
\(8 x-6 y=-11\)
\(\mathrm{a}^2=16 \quad \mathrm{~b}^2=64\)
\(y=\frac{4}{3} x \pm \sqrt{16 \times \frac{16}{9}-64}\)
\(y=\frac{4}{3} x \pm \sqrt{\frac{256-576}{9}}\)
\(y=\frac{4}{3} x \pm \sqrt{\frac{-320}{9}}\), an imaginary value of the constant
term.
So, Tangent does not exist.
GUJCET-2007
Hyperbola
120823
The curve for which the length of the normal is equal to the length of the radius vector, are
1 only circles
2 only rectangular hyperbolas
3 either circles or rectangular hyperbolas
4 None of the above
Explanation:
C We have-
length of normal \(=\) radius vector.
\(\mathrm{y} \sqrt{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2}=\sqrt{\mathrm{x}^2+\mathrm{y}^2}\) \(\mathrm{y}^2\left\{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right\}=\mathrm{x}^2+\mathrm{y}^2\)
\(y^2+y^2\left(\frac{d y}{d x}\right)^2=x^2+y^2\)
\(y^2\left(\frac{d y}{d x}\right)^2=x^2\)
\(x= \pm y \frac{d y}{d x}\)
Here, we have-
\(x=y \frac{d y}{d x} \quad \text { or } \quad x=-y \frac{d y}{d x}\)
Now,
\(\text { If } x=y \frac{d y}{d x}\)
\(x d x-y d y=0\)
\(x^2-y^2=c_1\)
and,
\(x=-y \frac{d y}{d x}\)
\(x d x+y d y=0\)
\(x^2+y^2=c_2\)Clearly, \(x^2-y^2=c_1\) represents a rectangular hyperbola and \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{c}_2\) represents circle.
Manipal UGET-2013
Hyperbola
120822
The line \(x \cos \alpha+y \sin \alpha=p\) touches the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), if:
A The line \(\mathrm{x} \cos \propto+\mathrm{y} \sin \propto=\mathrm{p}\) touches the hyperbola,
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { if }\)
\(a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2\)
