120248
If a normal chord at a point \(t(\neq 0)\) on the parabola \(y^2=9 x\) subtends a right angle at its vertex, then \(t=\)
1 \(\sqrt{3}\)
2 \(\sqrt{5}\)
3 \(\pm \sqrt{3}\)
4 \(\pm \sqrt{2}\)
Explanation:
D Equation of given parabola is
\(y^2=9 x\)
Equation o normal chord at point \(t(\neq 0)\) on the parabola (i) is -
\(\mathrm{Tx}+\mathrm{y}=2\left(\frac{9}{4}\right) \mathrm{t}+\frac{9}{4} \mathrm{t}^3\)
The chord (ii) subtends a right angle at vertex of parabola \(\mathrm{V}(0,0)\) so first homogenise the parabola (i) with the help of line (ii), we get -
\(y^2-9 x\left(\frac{t x+y}{\frac{9}{2} t+\frac{9}{4} t^3}\right)=0\)
The sum of the coefficients of \(x^2\) and \(y^2\) terms should be zero.
So,
\(1-\frac{9 t}{\frac{9}{2} t+\frac{9}{4} t^3}=0\)
\(1-\frac{1}{\frac{1}{2}+\frac{t^2}{4}}=0\)
\(\frac{1}{2}+\frac{t^2}{4}=0\)
\(t^2=2\)
\(t= \pm \sqrt{2}\)
AP EAMCET-20.04.2019
Parabola
120249
The locus of the points intersections of perpendicular normal's to the parabola \(y^2=\) 4ax, is
1 \(\mathrm{y}^2-2 \mathrm{ax}+2 \mathrm{a}^2=0\)
2 \(y^2+a x+2 a^2=0\)
3 \(\mathrm{y}^2-\mathrm{ax}+2 \mathrm{a}^2=0\)
4 \(y^2-a x+3 a^2=0\)
Explanation:
D Given, parabola \(-\mathrm{y}^2=4 \mathrm{ax}\)
Let the equation contain slope \(\mathrm{m}\) to the parabola is \(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3\) which passes through point \((\mathrm{h}, \mathrm{k})\) Then,
\(\mathrm{k}=\mathrm{mh}-2 \mathrm{am}-\mathrm{am}^3\)
Let the roots be \(\mathrm{m}_1, \mathrm{~m}_2\) and \(\mathrm{m}_3\) then
\(\mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=\frac{-\mathrm{k}}{\mathrm{a}}\)
Due to perpendicular normal \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\mathrm{m}_3=\frac{\mathrm{k}}{\mathrm{a}} \Rightarrow \mathrm{k}=\frac{\mathrm{k}}{\mathrm{a}} \mathrm{h}-\frac{2 \mathrm{ak}}{\mathrm{a}}-\mathrm{a}\left(\frac{\mathrm{k}}{\mathrm{a}}\right)^3\)
\(1=\frac{\mathrm{h}}{\mathrm{a}}-2-\frac{\mathrm{k}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{k}^2}{\mathrm{a}^2}=\frac{\mathrm{h}}{\mathrm{a}}-3 \Rightarrow \mathrm{k}^2=\frac{\mathrm{a}^2 \mathrm{~h}}{\mathrm{a}}-3 \mathrm{a}^2\)
\(\mathrm{k}^2=\mathrm{ah}-3 \mathrm{a}^2\)
\(=\mathrm{a}(\mathrm{h}-3 \mathrm{a})\)
Now, taking locus \((\mathrm{h}, \mathrm{k})\) we can get -
\(y^2-a x+3 a^2=0\)
AP EAMCET-20.04.2019
Parabola
120250
If \(a x+b y=1\) is a normal to the parabola \(y^2=\) \(4 \mathrm{px}\) then the condition is
C Given, \(\mathrm{ax}+\mathrm{by}=1\) is normal to \(\mathrm{y}^2=4 \mathrm{px}\) then,
\(\mathrm{y}=\frac{-\mathrm{a}}{\mathrm{b}} \mathrm{x}+\frac{1}{\mathrm{~b}}\)
Slope, \(\mathrm{m}=\frac{-\mathrm{a}}{\mathrm{b}}\), constant, \(\mathrm{c}=\frac{1}{\mathrm{~b}}\)
Now,
\(c=-2 m \left(\frac{\text { coefficient of } \mathrm{x}}{4}\right)-\left(\frac{\text { cofficient of } \mathrm{x}}{4}\right) \mathrm{m}^3\)
\(\frac{1}{\mathrm{~b}} =-2 \times\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)(\mathrm{p})-\mathrm{p}\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)^3\)
\(\frac{1}{\mathrm{~b}} =\frac{2 \mathrm{ap}}{\mathrm{b}}+\frac{\mathrm{pa}^3}{\mathrm{~b}^3}\)
\(\mathrm{~b}^2 =2 \mathrm{pab}^2+\mathrm{pa}^3\)
\(\mathrm{pa}^3 =\mathrm{b}^2-2 \mathrm{pab}^2\)
AP EAMCET-04.07.2021
Parabola
120251
If \(P\) and the origin are the points of intersection of the parabolas \(y^2=32 x\) and \(2 x^2=27 y\); and if \(\theta\) is the acute angle between these curves at \(P\), then \(5 \sqrt{\tan \theta}=\)
120252
The angle between the tangents drawn from the point \((1,4)\) to the parabola \(y^2=4 x\), is:
1 \(\pi / 2\)
2 \(\pi / 6\)
3 \(\pi / 4\)
4 \(\pi / 3\)
Explanation:
D We know that, equation of tangent on \(y^2=4 x\) is \(\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
As it passes through \((1,4)\) then,
\(4=m+\frac{1}{m}\)
\(m^2-4 m+1=0\)
Let, the roots \(\mathrm{m}_1\) and \(\mathrm{m}_2\)
\(\mathrm{m}_1+\mathrm{m}_2=4\)
\(\mathrm{~m}_1 \times \mathrm{m}_2=+1\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=16-4 \times 1\)
\(\left|\mathrm{~m}_1-\mathrm{m}_2\right|=\sqrt{12}=2 \sqrt{3}\)
\(\therefore \quad \tan \theta=\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \quad \frac{2 \sqrt{3}}{1+1}=\frac{2 \sqrt{3}}{2}=\sqrt{3}\)
\(\tan \theta=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}\)
120248
If a normal chord at a point \(t(\neq 0)\) on the parabola \(y^2=9 x\) subtends a right angle at its vertex, then \(t=\)
1 \(\sqrt{3}\)
2 \(\sqrt{5}\)
3 \(\pm \sqrt{3}\)
4 \(\pm \sqrt{2}\)
Explanation:
D Equation of given parabola is
\(y^2=9 x\)
Equation o normal chord at point \(t(\neq 0)\) on the parabola (i) is -
\(\mathrm{Tx}+\mathrm{y}=2\left(\frac{9}{4}\right) \mathrm{t}+\frac{9}{4} \mathrm{t}^3\)
The chord (ii) subtends a right angle at vertex of parabola \(\mathrm{V}(0,0)\) so first homogenise the parabola (i) with the help of line (ii), we get -
\(y^2-9 x\left(\frac{t x+y}{\frac{9}{2} t+\frac{9}{4} t^3}\right)=0\)
The sum of the coefficients of \(x^2\) and \(y^2\) terms should be zero.
So,
\(1-\frac{9 t}{\frac{9}{2} t+\frac{9}{4} t^3}=0\)
\(1-\frac{1}{\frac{1}{2}+\frac{t^2}{4}}=0\)
\(\frac{1}{2}+\frac{t^2}{4}=0\)
\(t^2=2\)
\(t= \pm \sqrt{2}\)
AP EAMCET-20.04.2019
Parabola
120249
The locus of the points intersections of perpendicular normal's to the parabola \(y^2=\) 4ax, is
1 \(\mathrm{y}^2-2 \mathrm{ax}+2 \mathrm{a}^2=0\)
2 \(y^2+a x+2 a^2=0\)
3 \(\mathrm{y}^2-\mathrm{ax}+2 \mathrm{a}^2=0\)
4 \(y^2-a x+3 a^2=0\)
Explanation:
D Given, parabola \(-\mathrm{y}^2=4 \mathrm{ax}\)
Let the equation contain slope \(\mathrm{m}\) to the parabola is \(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3\) which passes through point \((\mathrm{h}, \mathrm{k})\) Then,
\(\mathrm{k}=\mathrm{mh}-2 \mathrm{am}-\mathrm{am}^3\)
Let the roots be \(\mathrm{m}_1, \mathrm{~m}_2\) and \(\mathrm{m}_3\) then
\(\mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=\frac{-\mathrm{k}}{\mathrm{a}}\)
Due to perpendicular normal \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\mathrm{m}_3=\frac{\mathrm{k}}{\mathrm{a}} \Rightarrow \mathrm{k}=\frac{\mathrm{k}}{\mathrm{a}} \mathrm{h}-\frac{2 \mathrm{ak}}{\mathrm{a}}-\mathrm{a}\left(\frac{\mathrm{k}}{\mathrm{a}}\right)^3\)
\(1=\frac{\mathrm{h}}{\mathrm{a}}-2-\frac{\mathrm{k}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{k}^2}{\mathrm{a}^2}=\frac{\mathrm{h}}{\mathrm{a}}-3 \Rightarrow \mathrm{k}^2=\frac{\mathrm{a}^2 \mathrm{~h}}{\mathrm{a}}-3 \mathrm{a}^2\)
\(\mathrm{k}^2=\mathrm{ah}-3 \mathrm{a}^2\)
\(=\mathrm{a}(\mathrm{h}-3 \mathrm{a})\)
Now, taking locus \((\mathrm{h}, \mathrm{k})\) we can get -
\(y^2-a x+3 a^2=0\)
AP EAMCET-20.04.2019
Parabola
120250
If \(a x+b y=1\) is a normal to the parabola \(y^2=\) \(4 \mathrm{px}\) then the condition is
C Given, \(\mathrm{ax}+\mathrm{by}=1\) is normal to \(\mathrm{y}^2=4 \mathrm{px}\) then,
\(\mathrm{y}=\frac{-\mathrm{a}}{\mathrm{b}} \mathrm{x}+\frac{1}{\mathrm{~b}}\)
Slope, \(\mathrm{m}=\frac{-\mathrm{a}}{\mathrm{b}}\), constant, \(\mathrm{c}=\frac{1}{\mathrm{~b}}\)
Now,
\(c=-2 m \left(\frac{\text { coefficient of } \mathrm{x}}{4}\right)-\left(\frac{\text { cofficient of } \mathrm{x}}{4}\right) \mathrm{m}^3\)
\(\frac{1}{\mathrm{~b}} =-2 \times\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)(\mathrm{p})-\mathrm{p}\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)^3\)
\(\frac{1}{\mathrm{~b}} =\frac{2 \mathrm{ap}}{\mathrm{b}}+\frac{\mathrm{pa}^3}{\mathrm{~b}^3}\)
\(\mathrm{~b}^2 =2 \mathrm{pab}^2+\mathrm{pa}^3\)
\(\mathrm{pa}^3 =\mathrm{b}^2-2 \mathrm{pab}^2\)
AP EAMCET-04.07.2021
Parabola
120251
If \(P\) and the origin are the points of intersection of the parabolas \(y^2=32 x\) and \(2 x^2=27 y\); and if \(\theta\) is the acute angle between these curves at \(P\), then \(5 \sqrt{\tan \theta}=\)
120252
The angle between the tangents drawn from the point \((1,4)\) to the parabola \(y^2=4 x\), is:
1 \(\pi / 2\)
2 \(\pi / 6\)
3 \(\pi / 4\)
4 \(\pi / 3\)
Explanation:
D We know that, equation of tangent on \(y^2=4 x\) is \(\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
As it passes through \((1,4)\) then,
\(4=m+\frac{1}{m}\)
\(m^2-4 m+1=0\)
Let, the roots \(\mathrm{m}_1\) and \(\mathrm{m}_2\)
\(\mathrm{m}_1+\mathrm{m}_2=4\)
\(\mathrm{~m}_1 \times \mathrm{m}_2=+1\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=16-4 \times 1\)
\(\left|\mathrm{~m}_1-\mathrm{m}_2\right|=\sqrt{12}=2 \sqrt{3}\)
\(\therefore \quad \tan \theta=\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \quad \frac{2 \sqrt{3}}{1+1}=\frac{2 \sqrt{3}}{2}=\sqrt{3}\)
\(\tan \theta=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}\)
120248
If a normal chord at a point \(t(\neq 0)\) on the parabola \(y^2=9 x\) subtends a right angle at its vertex, then \(t=\)
1 \(\sqrt{3}\)
2 \(\sqrt{5}\)
3 \(\pm \sqrt{3}\)
4 \(\pm \sqrt{2}\)
Explanation:
D Equation of given parabola is
\(y^2=9 x\)
Equation o normal chord at point \(t(\neq 0)\) on the parabola (i) is -
\(\mathrm{Tx}+\mathrm{y}=2\left(\frac{9}{4}\right) \mathrm{t}+\frac{9}{4} \mathrm{t}^3\)
The chord (ii) subtends a right angle at vertex of parabola \(\mathrm{V}(0,0)\) so first homogenise the parabola (i) with the help of line (ii), we get -
\(y^2-9 x\left(\frac{t x+y}{\frac{9}{2} t+\frac{9}{4} t^3}\right)=0\)
The sum of the coefficients of \(x^2\) and \(y^2\) terms should be zero.
So,
\(1-\frac{9 t}{\frac{9}{2} t+\frac{9}{4} t^3}=0\)
\(1-\frac{1}{\frac{1}{2}+\frac{t^2}{4}}=0\)
\(\frac{1}{2}+\frac{t^2}{4}=0\)
\(t^2=2\)
\(t= \pm \sqrt{2}\)
AP EAMCET-20.04.2019
Parabola
120249
The locus of the points intersections of perpendicular normal's to the parabola \(y^2=\) 4ax, is
1 \(\mathrm{y}^2-2 \mathrm{ax}+2 \mathrm{a}^2=0\)
2 \(y^2+a x+2 a^2=0\)
3 \(\mathrm{y}^2-\mathrm{ax}+2 \mathrm{a}^2=0\)
4 \(y^2-a x+3 a^2=0\)
Explanation:
D Given, parabola \(-\mathrm{y}^2=4 \mathrm{ax}\)
Let the equation contain slope \(\mathrm{m}\) to the parabola is \(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3\) which passes through point \((\mathrm{h}, \mathrm{k})\) Then,
\(\mathrm{k}=\mathrm{mh}-2 \mathrm{am}-\mathrm{am}^3\)
Let the roots be \(\mathrm{m}_1, \mathrm{~m}_2\) and \(\mathrm{m}_3\) then
\(\mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=\frac{-\mathrm{k}}{\mathrm{a}}\)
Due to perpendicular normal \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\mathrm{m}_3=\frac{\mathrm{k}}{\mathrm{a}} \Rightarrow \mathrm{k}=\frac{\mathrm{k}}{\mathrm{a}} \mathrm{h}-\frac{2 \mathrm{ak}}{\mathrm{a}}-\mathrm{a}\left(\frac{\mathrm{k}}{\mathrm{a}}\right)^3\)
\(1=\frac{\mathrm{h}}{\mathrm{a}}-2-\frac{\mathrm{k}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{k}^2}{\mathrm{a}^2}=\frac{\mathrm{h}}{\mathrm{a}}-3 \Rightarrow \mathrm{k}^2=\frac{\mathrm{a}^2 \mathrm{~h}}{\mathrm{a}}-3 \mathrm{a}^2\)
\(\mathrm{k}^2=\mathrm{ah}-3 \mathrm{a}^2\)
\(=\mathrm{a}(\mathrm{h}-3 \mathrm{a})\)
Now, taking locus \((\mathrm{h}, \mathrm{k})\) we can get -
\(y^2-a x+3 a^2=0\)
AP EAMCET-20.04.2019
Parabola
120250
If \(a x+b y=1\) is a normal to the parabola \(y^2=\) \(4 \mathrm{px}\) then the condition is
C Given, \(\mathrm{ax}+\mathrm{by}=1\) is normal to \(\mathrm{y}^2=4 \mathrm{px}\) then,
\(\mathrm{y}=\frac{-\mathrm{a}}{\mathrm{b}} \mathrm{x}+\frac{1}{\mathrm{~b}}\)
Slope, \(\mathrm{m}=\frac{-\mathrm{a}}{\mathrm{b}}\), constant, \(\mathrm{c}=\frac{1}{\mathrm{~b}}\)
Now,
\(c=-2 m \left(\frac{\text { coefficient of } \mathrm{x}}{4}\right)-\left(\frac{\text { cofficient of } \mathrm{x}}{4}\right) \mathrm{m}^3\)
\(\frac{1}{\mathrm{~b}} =-2 \times\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)(\mathrm{p})-\mathrm{p}\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)^3\)
\(\frac{1}{\mathrm{~b}} =\frac{2 \mathrm{ap}}{\mathrm{b}}+\frac{\mathrm{pa}^3}{\mathrm{~b}^3}\)
\(\mathrm{~b}^2 =2 \mathrm{pab}^2+\mathrm{pa}^3\)
\(\mathrm{pa}^3 =\mathrm{b}^2-2 \mathrm{pab}^2\)
AP EAMCET-04.07.2021
Parabola
120251
If \(P\) and the origin are the points of intersection of the parabolas \(y^2=32 x\) and \(2 x^2=27 y\); and if \(\theta\) is the acute angle between these curves at \(P\), then \(5 \sqrt{\tan \theta}=\)
120252
The angle between the tangents drawn from the point \((1,4)\) to the parabola \(y^2=4 x\), is:
1 \(\pi / 2\)
2 \(\pi / 6\)
3 \(\pi / 4\)
4 \(\pi / 3\)
Explanation:
D We know that, equation of tangent on \(y^2=4 x\) is \(\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
As it passes through \((1,4)\) then,
\(4=m+\frac{1}{m}\)
\(m^2-4 m+1=0\)
Let, the roots \(\mathrm{m}_1\) and \(\mathrm{m}_2\)
\(\mathrm{m}_1+\mathrm{m}_2=4\)
\(\mathrm{~m}_1 \times \mathrm{m}_2=+1\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=16-4 \times 1\)
\(\left|\mathrm{~m}_1-\mathrm{m}_2\right|=\sqrt{12}=2 \sqrt{3}\)
\(\therefore \quad \tan \theta=\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \quad \frac{2 \sqrt{3}}{1+1}=\frac{2 \sqrt{3}}{2}=\sqrt{3}\)
\(\tan \theta=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}\)
120248
If a normal chord at a point \(t(\neq 0)\) on the parabola \(y^2=9 x\) subtends a right angle at its vertex, then \(t=\)
1 \(\sqrt{3}\)
2 \(\sqrt{5}\)
3 \(\pm \sqrt{3}\)
4 \(\pm \sqrt{2}\)
Explanation:
D Equation of given parabola is
\(y^2=9 x\)
Equation o normal chord at point \(t(\neq 0)\) on the parabola (i) is -
\(\mathrm{Tx}+\mathrm{y}=2\left(\frac{9}{4}\right) \mathrm{t}+\frac{9}{4} \mathrm{t}^3\)
The chord (ii) subtends a right angle at vertex of parabola \(\mathrm{V}(0,0)\) so first homogenise the parabola (i) with the help of line (ii), we get -
\(y^2-9 x\left(\frac{t x+y}{\frac{9}{2} t+\frac{9}{4} t^3}\right)=0\)
The sum of the coefficients of \(x^2\) and \(y^2\) terms should be zero.
So,
\(1-\frac{9 t}{\frac{9}{2} t+\frac{9}{4} t^3}=0\)
\(1-\frac{1}{\frac{1}{2}+\frac{t^2}{4}}=0\)
\(\frac{1}{2}+\frac{t^2}{4}=0\)
\(t^2=2\)
\(t= \pm \sqrt{2}\)
AP EAMCET-20.04.2019
Parabola
120249
The locus of the points intersections of perpendicular normal's to the parabola \(y^2=\) 4ax, is
1 \(\mathrm{y}^2-2 \mathrm{ax}+2 \mathrm{a}^2=0\)
2 \(y^2+a x+2 a^2=0\)
3 \(\mathrm{y}^2-\mathrm{ax}+2 \mathrm{a}^2=0\)
4 \(y^2-a x+3 a^2=0\)
Explanation:
D Given, parabola \(-\mathrm{y}^2=4 \mathrm{ax}\)
Let the equation contain slope \(\mathrm{m}\) to the parabola is \(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3\) which passes through point \((\mathrm{h}, \mathrm{k})\) Then,
\(\mathrm{k}=\mathrm{mh}-2 \mathrm{am}-\mathrm{am}^3\)
Let the roots be \(\mathrm{m}_1, \mathrm{~m}_2\) and \(\mathrm{m}_3\) then
\(\mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=\frac{-\mathrm{k}}{\mathrm{a}}\)
Due to perpendicular normal \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\mathrm{m}_3=\frac{\mathrm{k}}{\mathrm{a}} \Rightarrow \mathrm{k}=\frac{\mathrm{k}}{\mathrm{a}} \mathrm{h}-\frac{2 \mathrm{ak}}{\mathrm{a}}-\mathrm{a}\left(\frac{\mathrm{k}}{\mathrm{a}}\right)^3\)
\(1=\frac{\mathrm{h}}{\mathrm{a}}-2-\frac{\mathrm{k}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{k}^2}{\mathrm{a}^2}=\frac{\mathrm{h}}{\mathrm{a}}-3 \Rightarrow \mathrm{k}^2=\frac{\mathrm{a}^2 \mathrm{~h}}{\mathrm{a}}-3 \mathrm{a}^2\)
\(\mathrm{k}^2=\mathrm{ah}-3 \mathrm{a}^2\)
\(=\mathrm{a}(\mathrm{h}-3 \mathrm{a})\)
Now, taking locus \((\mathrm{h}, \mathrm{k})\) we can get -
\(y^2-a x+3 a^2=0\)
AP EAMCET-20.04.2019
Parabola
120250
If \(a x+b y=1\) is a normal to the parabola \(y^2=\) \(4 \mathrm{px}\) then the condition is
C Given, \(\mathrm{ax}+\mathrm{by}=1\) is normal to \(\mathrm{y}^2=4 \mathrm{px}\) then,
\(\mathrm{y}=\frac{-\mathrm{a}}{\mathrm{b}} \mathrm{x}+\frac{1}{\mathrm{~b}}\)
Slope, \(\mathrm{m}=\frac{-\mathrm{a}}{\mathrm{b}}\), constant, \(\mathrm{c}=\frac{1}{\mathrm{~b}}\)
Now,
\(c=-2 m \left(\frac{\text { coefficient of } \mathrm{x}}{4}\right)-\left(\frac{\text { cofficient of } \mathrm{x}}{4}\right) \mathrm{m}^3\)
\(\frac{1}{\mathrm{~b}} =-2 \times\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)(\mathrm{p})-\mathrm{p}\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)^3\)
\(\frac{1}{\mathrm{~b}} =\frac{2 \mathrm{ap}}{\mathrm{b}}+\frac{\mathrm{pa}^3}{\mathrm{~b}^3}\)
\(\mathrm{~b}^2 =2 \mathrm{pab}^2+\mathrm{pa}^3\)
\(\mathrm{pa}^3 =\mathrm{b}^2-2 \mathrm{pab}^2\)
AP EAMCET-04.07.2021
Parabola
120251
If \(P\) and the origin are the points of intersection of the parabolas \(y^2=32 x\) and \(2 x^2=27 y\); and if \(\theta\) is the acute angle between these curves at \(P\), then \(5 \sqrt{\tan \theta}=\)
120252
The angle between the tangents drawn from the point \((1,4)\) to the parabola \(y^2=4 x\), is:
1 \(\pi / 2\)
2 \(\pi / 6\)
3 \(\pi / 4\)
4 \(\pi / 3\)
Explanation:
D We know that, equation of tangent on \(y^2=4 x\) is \(\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
As it passes through \((1,4)\) then,
\(4=m+\frac{1}{m}\)
\(m^2-4 m+1=0\)
Let, the roots \(\mathrm{m}_1\) and \(\mathrm{m}_2\)
\(\mathrm{m}_1+\mathrm{m}_2=4\)
\(\mathrm{~m}_1 \times \mathrm{m}_2=+1\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=16-4 \times 1\)
\(\left|\mathrm{~m}_1-\mathrm{m}_2\right|=\sqrt{12}=2 \sqrt{3}\)
\(\therefore \quad \tan \theta=\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \quad \frac{2 \sqrt{3}}{1+1}=\frac{2 \sqrt{3}}{2}=\sqrt{3}\)
\(\tan \theta=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}\)
120248
If a normal chord at a point \(t(\neq 0)\) on the parabola \(y^2=9 x\) subtends a right angle at its vertex, then \(t=\)
1 \(\sqrt{3}\)
2 \(\sqrt{5}\)
3 \(\pm \sqrt{3}\)
4 \(\pm \sqrt{2}\)
Explanation:
D Equation of given parabola is
\(y^2=9 x\)
Equation o normal chord at point \(t(\neq 0)\) on the parabola (i) is -
\(\mathrm{Tx}+\mathrm{y}=2\left(\frac{9}{4}\right) \mathrm{t}+\frac{9}{4} \mathrm{t}^3\)
The chord (ii) subtends a right angle at vertex of parabola \(\mathrm{V}(0,0)\) so first homogenise the parabola (i) with the help of line (ii), we get -
\(y^2-9 x\left(\frac{t x+y}{\frac{9}{2} t+\frac{9}{4} t^3}\right)=0\)
The sum of the coefficients of \(x^2\) and \(y^2\) terms should be zero.
So,
\(1-\frac{9 t}{\frac{9}{2} t+\frac{9}{4} t^3}=0\)
\(1-\frac{1}{\frac{1}{2}+\frac{t^2}{4}}=0\)
\(\frac{1}{2}+\frac{t^2}{4}=0\)
\(t^2=2\)
\(t= \pm \sqrt{2}\)
AP EAMCET-20.04.2019
Parabola
120249
The locus of the points intersections of perpendicular normal's to the parabola \(y^2=\) 4ax, is
1 \(\mathrm{y}^2-2 \mathrm{ax}+2 \mathrm{a}^2=0\)
2 \(y^2+a x+2 a^2=0\)
3 \(\mathrm{y}^2-\mathrm{ax}+2 \mathrm{a}^2=0\)
4 \(y^2-a x+3 a^2=0\)
Explanation:
D Given, parabola \(-\mathrm{y}^2=4 \mathrm{ax}\)
Let the equation contain slope \(\mathrm{m}\) to the parabola is \(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3\) which passes through point \((\mathrm{h}, \mathrm{k})\) Then,
\(\mathrm{k}=\mathrm{mh}-2 \mathrm{am}-\mathrm{am}^3\)
Let the roots be \(\mathrm{m}_1, \mathrm{~m}_2\) and \(\mathrm{m}_3\) then
\(\mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=\frac{-\mathrm{k}}{\mathrm{a}}\)
Due to perpendicular normal \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
\(\mathrm{m}_3=\frac{\mathrm{k}}{\mathrm{a}} \Rightarrow \mathrm{k}=\frac{\mathrm{k}}{\mathrm{a}} \mathrm{h}-\frac{2 \mathrm{ak}}{\mathrm{a}}-\mathrm{a}\left(\frac{\mathrm{k}}{\mathrm{a}}\right)^3\)
\(1=\frac{\mathrm{h}}{\mathrm{a}}-2-\frac{\mathrm{k}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{k}^2}{\mathrm{a}^2}=\frac{\mathrm{h}}{\mathrm{a}}-3 \Rightarrow \mathrm{k}^2=\frac{\mathrm{a}^2 \mathrm{~h}}{\mathrm{a}}-3 \mathrm{a}^2\)
\(\mathrm{k}^2=\mathrm{ah}-3 \mathrm{a}^2\)
\(=\mathrm{a}(\mathrm{h}-3 \mathrm{a})\)
Now, taking locus \((\mathrm{h}, \mathrm{k})\) we can get -
\(y^2-a x+3 a^2=0\)
AP EAMCET-20.04.2019
Parabola
120250
If \(a x+b y=1\) is a normal to the parabola \(y^2=\) \(4 \mathrm{px}\) then the condition is
C Given, \(\mathrm{ax}+\mathrm{by}=1\) is normal to \(\mathrm{y}^2=4 \mathrm{px}\) then,
\(\mathrm{y}=\frac{-\mathrm{a}}{\mathrm{b}} \mathrm{x}+\frac{1}{\mathrm{~b}}\)
Slope, \(\mathrm{m}=\frac{-\mathrm{a}}{\mathrm{b}}\), constant, \(\mathrm{c}=\frac{1}{\mathrm{~b}}\)
Now,
\(c=-2 m \left(\frac{\text { coefficient of } \mathrm{x}}{4}\right)-\left(\frac{\text { cofficient of } \mathrm{x}}{4}\right) \mathrm{m}^3\)
\(\frac{1}{\mathrm{~b}} =-2 \times\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)(\mathrm{p})-\mathrm{p}\left(\frac{-\mathrm{a}}{\mathrm{b}}\right)^3\)
\(\frac{1}{\mathrm{~b}} =\frac{2 \mathrm{ap}}{\mathrm{b}}+\frac{\mathrm{pa}^3}{\mathrm{~b}^3}\)
\(\mathrm{~b}^2 =2 \mathrm{pab}^2+\mathrm{pa}^3\)
\(\mathrm{pa}^3 =\mathrm{b}^2-2 \mathrm{pab}^2\)
AP EAMCET-04.07.2021
Parabola
120251
If \(P\) and the origin are the points of intersection of the parabolas \(y^2=32 x\) and \(2 x^2=27 y\); and if \(\theta\) is the acute angle between these curves at \(P\), then \(5 \sqrt{\tan \theta}=\)
120252
The angle between the tangents drawn from the point \((1,4)\) to the parabola \(y^2=4 x\), is:
1 \(\pi / 2\)
2 \(\pi / 6\)
3 \(\pi / 4\)
4 \(\pi / 3\)
Explanation:
D We know that, equation of tangent on \(y^2=4 x\) is \(\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
As it passes through \((1,4)\) then,
\(4=m+\frac{1}{m}\)
\(m^2-4 m+1=0\)
Let, the roots \(\mathrm{m}_1\) and \(\mathrm{m}_2\)
\(\mathrm{m}_1+\mathrm{m}_2=4\)
\(\mathrm{~m}_1 \times \mathrm{m}_2=+1\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2\)
\(\left(\mathrm{~m}_1-\mathrm{m}_2\right)^2=16-4 \times 1\)
\(\left|\mathrm{~m}_1-\mathrm{m}_2\right|=\sqrt{12}=2 \sqrt{3}\)
\(\therefore \quad \tan \theta=\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \quad \frac{2 \sqrt{3}}{1+1}=\frac{2 \sqrt{3}}{2}=\sqrt{3}\)
\(\tan \theta=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}\)
