Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

Find the equation of tangent for ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1? \begin{array}{1 1}(a)\;\large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\normalsize=1\\(b)\;\large\frac{xx_1}{a^2}-\frac{yy_1}{b^2}\normalsize=1\\(c)\;\large\frac{xx_1}{a^3}+\frac{yy_1}{b^3}\normalsize=1\\(d)\;\text{None of these}\end{array} Can you answer this question? 1 Answer 0 votes Let us suppose a point P(x,y) at which we find the equation of tangent Let us find the slope at (x_1,y_1) Slope =\large\frac{2x}{a^2}+\frac{2yy'}{b^2}$$=0$
$y'=-\large\frac{-x_2\times b^2}{a^2\times y_1}$
Hence equation of tangent at $(x_1,y_1)$ is
$(y-y_1)=-\large\frac{-x_1\times b^2}{a^2\times y'}$$(x-x_1) \large\frac{yy_1}{b^2}-\frac{y_1^2}{b^2}=-\frac{xx_1}{a^2}+\frac{x_1^2}{a^2} \large\frac{xx_1}{a^2}+\frac{yy_1}{b^2}$$=1$
Hence (a) is the correct answer.
answered Feb 7, 2014

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}