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P Pankaj Sanodiya
As we know that if a point moves in a plane in such a way that its distance from two-point remain constant then the path is an ellipse. Now, According to the question, the distance between the point from where the sum of  the distance from a point is constant =  10 Now, the distance between the foci=8 Now, As we know the relation, Hence the equation of the ellipse is, Hence the path of...

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P Pankaj Sanodiya
Given, an equilateral triangle inscribed in parabola with the equation. The one coordinate of the triangle is A(0,0). Now, let the other two coordinates of the triangle are   and  Now, Since the triangle is equilateral, The coordinates of the points of the equilateral triangle are, So, the side of the triangle is   

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P Pankaj Sanodiya
Given the parabola, Comparing this equation with , we get Now, As we know the coordinates of ends of latus rectum are: So, the coordinates of latus rectum are, Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3) Widht of the triangle = 2*6=12 Height of the triangle = 3 So The area =  Hence the required area is 18 unit square.

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P Pankaj Sanodiya
Let  be the angle that rod makes with the ground, Now, at a point 3 cm from the end, At the point touching the ground Now, As we know the trigonometric identity, Hence the equation is,

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P Pankaj Sanodiya
The equation of the semi-ellipse will be of the form  Now, According to the question, the length of major axis = 2a = 8   The length of the semimajor axis =2 Hence the equation will be, Now, at point 1.5 cm from the end, the x coordinate is 4-1.5 = 2.5 So, the height at this point is  Hence the height of the required point is 1.56 m.

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P Pankaj Sanodiya
Given, The width of the parabolic cable = 100m The length of the shorter supportive wire attached =  6m The length of the longer supportive wire attached = 30m Since the rope opens towards upwards, the equation will be of the form  Now if we consider origin at the centre of the rope, the equation of the curve will pass through points, (50,30-6)=(50,24) Hence the equation of the parabola...

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P Pankaj Sanodiya
Since the Axis of the parabola is vertical, Let the equation of the parabola be,  it can be seen that this curve will pass through the point (5/2, 10) if we assume origin at the bottom end of the parabolic arch. So, Hence, the equation of the parabola is  Now, when y = 2 the value of x will be  Hence the width of the arch at this height is 

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P Pankaj Sanodiya
Le the parabolic reflector opens towards the right. So the equation of parabolic reflector will be, Now, Since this curve will pass through the point (5,10) if we assume origin at the optical centre, So Hence, The focus of the parabola is, . Alternative Method, As we know on any concave curve Hence, Focus  . Hence the focus is 5 cm right to the optical centre.

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P Pankaj Sanodiya
Given, in a hyperbola,  Foci , passing through (2,3) Since foci of the hyperbola are in Y-axis, the equation of the hyperbola will be of the form ; By comparing standard parameter (foci) with the given one, we get Now As we know, in a hyperbola  Now As the hyperbola passes through the point (2,3) Solving Equation (1) and (2) Now, as we know that in a hyperbola  is always greater...

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P Pankaj Sanodiya
Given, in a hyperbola vertices (± 7,0), And   Here, Vertices is  on the X-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (Vertices and eccentricity) with the given one, we get  and  From here, Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ;

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P Pankaj Sanodiya
Given, in a hyperbola Foci (± 4, 0), the latus rectum is of length 12 Here,  focii are on the X-axis so, the standard equation of the Hyperbola will be ; By comparing standard parameter (length of latus rectum and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Since  can never be negative, Hence, The Equation of the hyperbola is ;

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P Pankaj Sanodiya
Given, in a hyperbola Foci , the latus rectum is of length 8. Here,  focii are on the X-axis so, the standard equation of the Hyperbola will be ; By comparing standard parameter (length of latus rectum and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Since  can never be negative, Hence, The Equation of the hyperbola is ;

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P Pankaj Sanodiya
Given, in a hyperbola Foci (0, ±13), the conjugate axis is of length 24. Here, focii are on the Y-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (length of conjugate axis and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ; .

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P Pankaj Sanodiya
Given, in a hyperbola Foci (± 5, 0), the transverse axis is of length 8. Here,  focii are on the X-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (transverse axis length and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ;

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P Pankaj Sanodiya
Given, in a hyperbola  Vertices (0, ± 3), foci (0, ± 5) Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (Vertices and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ; .

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P Pankaj Sanodiya
Given, in a hyperbola Vertices (0, ± 5), foci (0, ± 8) Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (Vertices and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ; .

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P Pankaj Sanodiya
Given, in a hyperbola Vertices (± 2, 0), foci (± 3, 0) Here, Vertices and focii are on the X-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (Vertices and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence,The Equation of the hyperbola is ;

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P Pankaj Sanodiya
Given a Hyperbola equation, Can also be written as   Comparing this equation with the standard equation of the hyperbola: We get,  and  Now, As we know the relation in a hyperbola, Therefore, Coordinates of the foci: The Coordinates of vertices: The Eccentricity: The Length of the latus rectum :

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P Pankaj Sanodiya
Given a Hyperbola equation, Can also be written as  Comparing this equation with the standard equation of the hyperbola: We get,    and  Now, As we know the relation in a hyperbola, Here as we can see from the equation that the axis of the hyperbola is Y-axis. So,  Coordinates of the foci: The Coordinates of vertices: The Eccentricity: The Length of the latus rectum :

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P Pankaj Sanodiya
Given a Hyperbola equation, Can also be written as  Comparing this equation with the standard equation of the hyperbola: We get,  and  Now, As we know the relation in a hyperbola, Therefore, Coordinates of the foci: The Coordinates of vertices: The Eccentricity: The Length of the latus rectum :
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