タンジェント 2(x^2+y^2)^2=25(x^2-y^2), (3,1)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

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log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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tangent of 2(x²+y²)²=25(x²−y²), (3, 1)

Solution

y=−913 x+4013

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Find the slope of 2(x²+y²)²=25(x²−y²): x(25−4x²−4y²)y(4x²+4y²+25)

Compute the slope at (3, 1): m=−913

Find the line with slope m=−913 and passing through (3, 1): y=−913 x+4013

y=−913 x+4013

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

タンジェント 2(x^2+y^2)^2=25(x^2-y^2), (3,1)

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