हिंदी

Show that the Curves X 2 a 2 + λ 1 + Y 2 B 2 + λ 1 = 1 and X 2 a 2 + λ 2 + Y 2 B 2 + λ 2 = 1 Intersect at Right Angles ? - Mathematics

Advertisements
Advertisements

प्रश्न

Show that the curves \[\frac{x^2}{a^2 + \lambda_1} + \frac{y^2}{b^2 + \lambda_1} = 1 \text { and } \frac{x^2}{a^2 + \lambda_2} + \frac{y^2}{b^2 + \lambda_2} = 1\] intersect at right angles ?

Advertisements

उत्तर

\[\text { We have, } \frac{x^2}{a^2 + \lambda_1} + \frac{y^2}{b^2 + \lambda_1} = 1 . . . \left( 1 \right)\]

\[\text { and } \frac{x^2}{a^2 + \lambda_2} + \frac{y^2}{b^2 + \lambda_2} = 1 . . . \left( 2 \right)\]

\[\text { Now we can find the slope of both the curve by differentiating w . r . t x }\]

\[ \Rightarrow \frac{2x}{a^2 + \lambda_1} + \frac{2y\frac{dy}{dx}}{b^2 + \lambda_1} = 0 \text { and } \frac{2x}{a^2 + \lambda_2} + \frac{2y\frac{dy}{dx}}{b^2 + \lambda_2} = 0\]

\[ \Rightarrow \frac{dy}{dx} = - \frac{x}{y} \times \frac{b^2 + \lambda_1}{a^2 + \lambda_1} \text { and } \frac{dy}{dx} = - \frac{x}{y} \times \frac{b^2 + \lambda_2}{a^2 + \lambda_2}\]

\[ \Rightarrow m_1 = - \frac{x}{y} \times \frac{b^2 + \lambda_1}{a^2 + \lambda_1} \text { and } m_2 = - \frac{x}{y} \times \frac{b^2 + \lambda_2}{a^2 + \lambda_2}\]

\[\text { Subtracting} \left( 2 \right) \text { from } \left( 1 \right), \text { we get }, \]

\[ x^2 \left( \frac{1}{a^2 + \lambda_1} - \frac{1}{a^2 + \lambda_2} \right) + y^2 \left( \frac{1}{b^2 + \lambda_1} - \frac{1}{b^2 + \lambda_2} \right) = 0\]

\[ \Rightarrow \frac{x^2}{y^2} = \frac{\lambda_2 - \lambda_1}{\left( b^2 + \lambda_1 \right)\left( b^2 + \lambda_2 \right)} \times \frac{1}{\frac{\lambda_1 - \lambda_2}{\left( a^2 + \lambda_1 \right)\left( a^2 + \lambda_2 \right)}}\]

\[\text { Now,} \]

\[ m_1 \times \times m_2 = \frac{x^2}{y^2} \times \frac{b^2 + \lambda_1}{a^2 + \lambda_1} \times \frac{b^2 + \lambda_2}{a^2 + \lambda_2}\]

\[ = \frac{\lambda_2 - \lambda_1}{\left( b^2 + \lambda_1 \right)\left( b^2 + \lambda_2 \right)} \times \frac{\left( a^2 + \lambda_1 \right)\left( a^2 + \lambda_2 \right)}{\lambda_1 - \lambda_2} \times \frac{b^2 + \lambda_1}{a^2 + \lambda_1} \times \frac{b^2 + \lambda_2}{a^2 + \lambda_2}\]

\[ = - 1\]

\[\text { hence,} \left( 1 \right) \text { and } \left( 2 \right) \text { cuts orthogonally } . \]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 16: Tangents and Normals - Exercise 16.3 [पृष्ठ ४१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 16 Tangents and Normals
Exercise 16.3 | Q 9 | पृष्ठ ४१

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

Find the slope of the tangent to the curve y = (x -1)/(x - 2), x != 2 at x = 10.


Find the slope of the tangent to curve y = x3 − + 1 at the point whose x-coordinate is 2.


Find the equations of the tangent and normal to the given curves at the indicated points:

y = x2 at (0, 0)


Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]


Find the equation of the tangent to the curve `y = sqrt(3x-2)`  which is parallel to the line 4x − 2y + 5 = 0.

 

Find the equation of the normal to curve y2 = 4x at the point (1, 2).


Find the slope of the tangent and the normal to the following curve at the indicted point y = 2x2 + 3 sin x at x = 0 ?


Find the slope of the tangent and the normal to the following curve at the indicted point  xy = 6 at (1, 6) ?


Find the points on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45° with the x-axis ?


At what points on the circle x2 + y2 − 2x − 4y + 1 = 0, the tangent is parallel to x-axis?


Find the points on the curve y = 3x2 − 9x + 8 at which the tangents are equally inclined with the axes ?


At what points on the curve y = x2 − 4x + 5 is the tangent perpendicular to the line 2y + x = 7?


Find the equation of the tangent to the curve \[\sqrt{x} + \sqrt{y} = a\] at the point \[\left( \frac{a^2}{4}, \frac{a^2}{4} \right)\] ?


Find the equation of the tangent and the normal to the following curve at the indicated point  y = x2 at (0, 0) ?


Find the equation of the tangent and the normal to the following curve at the indicated point y = 2x2 − 3x − 1 at (1, −2) ?


Find the equation of the tangent and the normal to the following curve at the indicated points x = at2, y = 2at at t = 1 ?


Find an equation of normal line to the curve y = x3 + 2x + 6 which is parallel to the line x+ 14y + 4 = 0 ?


Find the equation of the tangent line to the curve y = x2 − 2x + 7 which perpendicular to the line 5y − 15x = 13. ?


Find the angle of intersection of the following curve  2y2 = x3 and y2 = 32x ?


Find the angle of intersection of the following curve  x2 + 4y2 = 8 and x2 − 2y2 = 2 ?


Show that the following curve intersect orthogonally at the indicated point y2 = 8x and 2x2 +  y2 = 10 at  \[\left( 1, 2\sqrt{2} \right)\] ?


If the straight line xcos \[\alpha\] +y sin \[\alpha\] = p touches the curve  \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\] then prove that a2cos2 \[\alpha\] \[-\] b2sin\[\alpha\] = p?


Write the slope of the normal to the curve \[y = \frac{1}{x}\]  at the point \[\left( 3, \frac{1}{3} \right)\] ?


The point on the curve y2 = x where tangent makes 45° angle with x-axis is ______________ .


Find the angle of intersection of the curves y2 = 4ax and x2 = 4by.


Find an angle θ, 0 < θ < `pi/2`, which increases twice as fast as its sine.


At (0, 0) the curve y = x3 + x


For which value of m is the line y = mx + 1 a tangent to the curve y2 = 4x?


The slope of the tangent to the curve x = a sin t, y = a{cot t + log(tan `"t"/2`)} at the point ‘t’ is ____________.


The line y = x + 1 is a tangent to the curve y2 = 4x at the point


The points at which the tangent passes through the origin for the curve y = 4x3 – 2x5 are


Let `y = f(x)` be the equation of the curve, then equation of normal is


The normal at the point (1, 1) on the curve `2y + x^2` = 3 is


If the curves y2 = 6x, 9x2 + by2 = 16, cut each other at right angles then the value of b is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×